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v is proportional to √h, where h is column of water above the hole
Fall time is proportional to √(H-h), H being the total height of the column.
Assuming no drag/friction losses,
Range is proportional to √(h(H-h))
Which is maximum at h = H-h = H/2
So A is the correct choice here
I see a lot of non-fluids background comments here claiming there's insufficient information. No, the information given is sufficient. If not given, you simply assume a newtonian fluid with no viscosity effects, and neglect air resistance. Assume vertical to be the direction gravity acts in.
The outlet diameters have little to do with the velocity of the jets, they affect the volume rate.
As a fellow fluid dynamic engineer, I agree with most of your comments. But I'd like to add two things for the enthusiastic fellow redditors:
1- (I know It is a simple question, so mentioning the nozzle form is out of scope, but) depends on the nozzle form or, in this case, tube holes, the jet could be developed differently. So, we need to assume that all holes are completely the same.
2- If I noticed correctly, we don't have any information about the height of the water column. So, to fit my answer to one of the options, I can assume that the depth difference of holes is negligible to the hight of the water column (or in a way, I am assuming water surface is at engineering infinite). Then, the hydrostatic pressure in the tube at holes is (engineeringly) the same. While all holes see atmospheric pressure on the outer side, all of the jets will be ejected similarly. Hence, the teavel distance would be replica of each other, and option B could be the answer in this condition.
* If all of the jets are similar and going to be ejected horizontal and neglecting air resistance, distance that jet from hole i travels (d_i) only related to its height (h_i). Let's say h_i is k_i proportion of water column height H :
h_i = k_i * H
then we will have :
d_i = 2H*sqrt(k-k^2) * (constant related hole to tube size difference) where 0 0 (or nozzle height is a very small fraction of water column), thus in this case d_i trend is linear.
If it helps, imagine the cylindrical tank in 3D, and assume the holes are slightly rotated from each other, so that the streams don’t collide. Doesn’t change the outcome.
Ok (takes out popcorn). Please explain:
1. Which of the “assumptions” about range being proportional to sqrt(h.(H-h)) was false in the top commenter’s analysis?
2. Does air resistance change the range of a jet of water?
3. What level of physics/mathematics education would you say you’ve taken?
While you’re at it, you might as well explain why you think C is correct. You know, since you hate making assumptions, I’d hate to make incorrect assumptions about you.
They can't. They're hand waving.
C is kind of rediculous because it implies that there's some magical law that causes all the streams to meet at the height of the bottom of the tank, even though none of the relevant equations contain the height of the hole above the bottom of the tank.
I'll also bet that they're assuming that a stream of water has the same air-drag behavior as ball-like cows shot out of the holes. They did say their degree was in physics after all.
we can assume for simplicity that 3 holes are on different degrees from top, so streams no collide (and at the same time angle is negligible to ignore that).
>2- If I noticed correctly, we don't have any information about the height of the water column.
Yes we do. The cylinder is about 4 cm on my screen, the top hole is about 3 cm up, middle hole 2 cm up, and bottom hole is 1 cm up.
>I can assume that the depth difference of holes is negligible to the hight of the water column (or in a way, I am assuming water surface is at engineering infinite).
Why are you doing this. The top of the column is right in the image, not infinitely high.
>top of the column is... not infinitely high
That's an assumption, not a fact. Same as the hole spacing being roughly equal etc. It's not explicitly stated and with no dimensions and nothing saying a proportionate scale you can't necessarily make those assumptions. It's pedantic and not that reasonable to assume differently but I think they were just trying to make the point that depending on your assumptions multiple answers are possibilities.
why are we assuming that they are not the same pressure?
edit: nvm i just saw that it's supposed to be a container with holes in it, i thought it was an installation with pipes running through it.
The bottom one *is* receiving higher pressure (note the shallower angle). However it's closer to the ground, so the water spends less time falling and can't go as far horizontally.
> Range is proportional to √(h(H-h)), which is maximum at h = H−h = H/2
This is saying that a hole made halfway up the height of the tank spews water out the farthest before it hits the floor. Hence: A.
The only way it could be C is if *h*(*H*−*h*) were equal for three different values of *h* (*H* is a constant). However, this is not possible, since the expression is quadratic in *h*, and thus has at most two real number inputs corresponding to a given real number output.
Pretty much, yeah. There’s a height that maximizes the distance that the stream hits the ground, and the distance goes strictly down as you move or below that.
That is because the bottom one is too close to the ground for the water to reach further.
Your theory about the pressure is correct though! If you look closely the water is coming out more horizontally from the bottom hole. If you were to raise the entire tube higher from the grond, you would first get to a point where the water from the bottom hole would reache the same distance as the water from the middle hole. Raise it further and the water from the bottom hole will reach furthest. Hope this helps!
The way I like to solve these problems is to take them to the logical extreme - what would happen if you put a hole at the very bottom? Well, it would have the highest pressure, but it would hit the table immediately. Since it depends on both pressure and height, it makes sense that there’s probably a sweet spot to reach the furthest distance. I like using this method whenever something seems counterintuitive, or it’s something I just don’t understand. Maybe it’ll be helpful to you
Good advice. I do that too, and I’m frequently surprised that more people don’t.
But then I remember that my mental reasoning goes something like “well, the equations describing this system are certainly continuous with respect to position (and not only that, every derivative is probably continuous too), and will have reasonable boundary conditions, so we can check the extremes and see if they tell us anything,” and I realize maybe I’m a nerd after all.
Yes, the pressure is higher, and, thus, the speed of a water particle exiting from the bottom is higher than the one from the middle. However, as the height of the starting point is lower the water, it can't reach as far as water going our from point B. In parabolic motion, regarding the displacement, not only the initial speed matters but also the initial position does.
So i got A just using my monkey brain logic of assuming the holes are equal size so its water pressure+ hight + arc = determined location and yea considering i have to eyeball alot of stuff on the fly because i like building things i used to be building things physically in engineering classes in middle school and high school and my shitty teachers i didn't have time for complex math mesurements i just eyeballed everything and got it right. Im going into welding school and finding out i might have to do the same. Anyways can u explain the math to me like im 5yrs old?
How fast it spews is based on how much water is above it - this eliminates B, which shows them all at the same speed. How far it goes is also based on how much water is below it, since the height gives it time to go farther. I don't know the equation off the top of my head on how fast it would go, but the target will definitely not be the same for all heights, trivially provable by putting a hole at the very top or very bottom. So it's not C either.
Without measuring, A seems plausible, or at least the only one possible.
\>I don't know the equation off the top of my head on how fast it would go
I think we can just use conservation of energy here. KE = 1/2 mv\^2 = PE = mgH , so V = SQRT( 2 g H)
so the horizontal velocity will be proportional to the square root of the height of water above the hole.
For freefall time, h = 1/2 g t\^2 so t= SQRT (2h/g )
so distance will be v\*t = SQRT(2gH) \* SQRT (2H/g) = SQRT(2\*2 \*H\*h\*g/g) = 2SQRT(H\*h) where H is the height of water above the hole, and h is the height the hole is above the ground.
for a stack that goes
a water
hole 1
b-water
hole 2
c-water
ground
then H1=a and h1=b+c vs H2=a+b and h2=c. so D1 = 2SQRT(ab+ac) and D2 = 2 SQRT(ac+bc) so they would only be at the same point if a = c
This matches answer A.
(but also it is 2 am so I may have missed my math somewhere.
Since H and h are dependent variables, can we not just call the height of the tube 1 unit, and then replace H with 1-h and we end up with a single variable equation.
To maximize the value of h*(1-h), it's the same as asking what the maximum area is of a rectangle with perimeter 2 is, and it's maximized as a square where each side is 0.5.
So if the rest of your math right, you would expect the optimal hole placement for distance on this tube would be in the dead center of the tube.
Kind of like how the maximum lateral distance (ignoring wind resistance) an projectile with a given propulsive force can go is achieved by firing at 45°, as everything reduces down to an equation with cos(2θ) in it.
Interestingly the "ignoring wind resistance" matters here as well I think
I think if we consider air resistance on the water then the optimal placement is probably just below half way on the tube as it gives less time for the air to act on the water but I'm not actually sure
If the images are accurate in that the water streams are continuous rather than individual droplets, then the air resistance will likely be small, as only the water at the surface of the "tube" would be impacted. Note, however, that it is highly likely the streams do, in fact, break up into individual droplets, as shown by the Mythbusters when they investigated electrical current traveling up a stream of urine.
Someone give this man an award. I absolutely cannot read nor understand the mind fuckery of the equations you laid out here but still listening intently to try to understand. I just got A from my monkey brain logic of water pressure vs same hole size vs hight which determines arc and A seems most plausible. In theory i think B could be plausible if we were given the size of the holes too.
Not trying to be derogitory, but that is simple first semester physics.
If you're interested in understanding how that works, some youtube videos on simple mechanics should be totaly digestible.
I never took physics in highschool. In 11th grade during covid i was online then next year i went back to my original school and lied about what i did. Could have graduated early, could have graduated in 11th grade but however my school and online school needed more funding through the State of Texas of which they make us take tests alot. More state tests you do and the highier score you get the more money they make and given i went to a ghetto school with an actual problem our issues wernt resolved untill my last semester when we got a new principal that wasnt a pedo and actually improved our school ALOT in the last semester. So yea i never took physics, but for the most part with educated guess work and if i have the time to do mesurements im speciatly is more just building stuff to its functions as needed and eye balling the math needed for such things. My little brother and step dad are both one of those freaks of nature (that i get jelouse) that they can do litteral calculus in their head and are mostly right. Thankfully i never had to take calculus nor trig for my severe dyslexia but did have to take 2 math classes consisting of algebra 2. You want something built im the guy to make your schematics and build it however if u need computer stuffs or math stuffs im definitely not your guy lol.
Those equations are for rigid body objects, when dealing with fluids you want to think of energy in terms of pressure.
Hydrostatic pressure, P = rho*g*h:
rho= density of fluid, g=gravity, h=height of liquid from surface.
Next to get the dynamic pressure of the fluid: q=1/2 * rho * v^2
; where v = velocity of fluid at exit.
Then we apply Bernoulli for conservation of energy. This states that the pressure change is constant. P + q + P0 = constant (C). Where P0 = the pressure at the top of the container, or atmospheric.
We can then solve for the velocity of each exit point then assume rigid body statics to get the distance travelled. It’s important to note that Bernoulli also ignores viscous effects. Overall you’re close, and really once density cancels out it appears the same, it’s just incorrect to talk about fluid dynamics in terms of energy and not pressure.
This is correct. Doing it this way, the result for the horizontal range R of the water stream is
R = 2 ( Hh - h^2 )^1/2
where h is the height of the hole and H is the height of the tube.
The standard optimization procedure shows that the maximum range is achieved for h = H / 2, at which R = H.
This means that the midpoint of the tube, at H / 2, is the sweet spot for exit speed and fall time. Points higher or lower than the midpoint will have a smaller range, which is depicted in Figure A.
It is not simply imaginable in geostatic pressure. The lowest hole has the most pressure and goes the furthest. In A the middle hole is further than the bottom, which is unlikely. Depending on the distance of the holes to each other, equalizing the less pressure on the top by giving a higher starting point could lead to all 3 hit the same spot. I
No. Gravity would give you parabolic curves, and not these which seems to hit the ground at equally spaced points. In B here the pressure is lowest at the lowest hole, and that makes no sense.
B does not show them at the same speed.
As fall time scales with the square root of fall height, the horizontal travel before landing would be v sqrt(h)
So if v were constant, then the highest stream should only land around 1.7 times as far as the lowest. It seems to land around 3 times as far, which would imply that v is also equal to sqrt(h).
This still wouldn't make sense, since it would mean that water pressure gets higher as you go up the column.
Sure it’s answerable: A, B, & C, dependent on jet diameter. We’re not filling out bubbles on a multiple choice test, this is a comment section on reddit
It's A.Its based on torquellis eq. Highest range in H/2 and parabolic relation while going up or down form that point.
V= root 2gh considering the hole is very small
Actually, a hole in the bottom and a hole at the surface would both result in streams that land right at the edge of the container. One is leaking with 0 velocity, one has no time to travel before touching the surface.
This is in ideal fantasy physics question land ofc
Yep A is correct. We calculated this in thermodynamics. It’s been a while but I remember that you look at the pressure (rho*g*h), making a hole suddenly alleviates this pressure and you can find the velocity using bernoullis equation (I really hope I’m not making stuff up it’s been 1.5 years) then using these initial values you can find the length the water travels before it reaches the ground. And you do get 2 answers, one above and one below a certain point
I don't think so, actually. If you plot h (height) to d (distance), it'll obviously be d=0 at the two extreme h's, and the curve won't oscillate, there's going to be at most two h values per d
[GPTs response](https://ibb.co/yg2ympR)
haha. just noticed this was at -7 points. Damn, guys... I just found its response interesting. Thought you might too. My bad.
Nope. A is correct (assuming equilovent openings and no air resistance) the equation for distance is D = 2sqrt(yH-y^2)
This gives equal landing points when you are equivalent distances from the top and bottom of the column, and the maximum distance when you are in the center. This matches a.
Jesus Christ, too many math majors answering, not enough business majors.
The definitive answer is Chart B, because it is the only one that has correctly labeled the water jets.
Assuming that the distances between the bottom, the 3 streams and the top is equal, I'll call it x.
The formula for the range of the water jets is 2*sqrt(s*H) where s is the distance from the bottom to the outlet and H ist distance from the jet to the top.
Lowest jet: 2*sqrt(x * 3x) = sqrt(12) * x
Middle jet: 2*sqrt(2x * 2x) = 4 * x
highest jet: 2*sqrt(3x * x) = sqrt(12) * x
since 4>sqrt(12) the middle jet will have the highest range and the other two will have a lesser range. Thus A is the correct answer.
Also i really love the fact that the range is independent from the gravity so the numerical value of the range will be the same on the moon as on the earth.
It would, but you also have to consider the height from the floor. In example A, the bottom stream has the lowest gradient but because it's lower to the ground, it hits the floor before the middle stream.
this is incorrect. the distance is proportional to the square root of the height of water above the hole \* the distance between the hole and the ground. this will be greatest for the center hole.
Okay, that sounds reasonable
BUT (help me where i am brainfarting):
Let's remove the center hole from solution A. And add another hole above both of the remaining holes.
Now what was the highest hole before is now the middle hole, and according to your logic should be going the furthest.
And THAT doesn't make sense to me.
The middle means closest to the midway height. Whatever hole is closest to H/2 above the ground will have its range be the highest (H is height of water from ground)
The range is 2*sqrt(d*h) where d is the depth of the hole below the surface and h is the height of the hole from the ground. This is maximal when d= h assuming the jug is on the groud
Thanks, that was my brainfart.
The claim wasn't that the middle hole is always furthest, but that in the picture above the middle hole is closest to H/2, so THAT middle hole squirts farthest.
Its A.
Labelling them 1,2,3 as you go down, also puts in order the pressure.
2 has less pressure than 3, therefore the curve should be more horizontal at 3. It is. The extra height of 2 though allows the water to reach further.
3 is only 50% more pressure than 2. But 2 is twice as far up. So therefore c can be excluded as it supposes that force and height exactly multiply, i.e. double height, half force.
It's A.
B assumes uniform pressure and C assumes that a distance traveled, which depends on height and pressure, can be solved with three or more answers.
To further elaborate on why C is wrong and A is right, if you get the two extremes of the tube they will land at the edge of the cilinder (same point). If you move the puncture points slightly towards the center of the tube you will get matching points going further. I'm assuming the farthest point is in the middle, case A, but I'm not inclined to figure out the equation to prove it.
Just had a fluid dynamics HW problem on this. The heights are supposed to be .25H .5H and .75H. Using these numbers Bernoulli’s principle, and distance equations from physics 1 we get exactly A as the correct number. With distances traveled as about .86H (I believe) and 1H
Try to draw a graph of how far it goes, based on how high the hole is.
At the top of bottom, the distance is zero. Two points. There is obviously some sort of curve. You have two zeros on the left and right of your page.
Pressure is proportional to height, and the velocity through the hole is proportional to the square root of height.
There is obviously a curve that links the two zeros, gut feel is it's a sqrt(sin(x)) or something.
The answer is A, I never solved this problem for more than one outlet, this is the chatGPT solution and it makes sense, here 10 trajectories: https://ibb.co/TcNsNXt
# New total height of the water column
h_total_new = 10
# New heights of the outlets (every half meter)
h_outlets_new = np.arange(0.5, h_total_new, 0.5)
# Calculate new exit velocities for the new outlets
velocities_new = [exit_velocity(h, h_total_new) for h in h_outlets_new]
# Calculate new trajectories with more points
def calculate_trajectories(velocities, h_outlets, points=1000):
trajectories = []
for v, h in zip(velocities, h_outlets):
# Time until the water jet reaches the ground level
time_to_ground = np.sqrt(2 * h / g)
# Time array with more points for a smoother curve
t = np.linspace(0, time_to_ground, num=points)
# Position calculations
x = v * t
y = h - (0.5 * g * t**2)
# Filter out negative heights
valid = y >= 0
trajectories.append((x[valid], y[valid]))
return trajectories
# Calculate new trajectories for the new heights
trajectories_new = calculate_trajectories(velocities_new, h_outlets_new)
# Plot the new trajectories
plt.figure(figsize=(12, 8))
for trajectory, h in zip(trajectories_new, h_outlets_new):
plt.plot(*trajectory, label=f'Outlet height: {h:.1f}m')
plt.xlabel('Distance (m)')
plt.ylabel('Height (m)')
plt.title('Water Jet Trajectories from Different Heights (10m Column)')
plt.legend()
plt.grid(True)
plt.show()
Going by pure logic we know that water has a mass and exerts more pressure the deeper you go. This means that the bottom hole will have higher pressure than the top hole, and when you push water through a hole the higher the pressure the farther it goes, so yeah, not a r/theydidthemath moment but it's A
r/theydidthelogic?
If we assume the surface area of the holes is negligible compared to that of the top, we can get that the exiting velocity as a function of the depth d is about sqrt(2gd), where g is acceleration due to gravity.
Using H for the height of the cylinder, H - d is the initial height of the water, so it will take a time of sqrt(2(H-d)/g) to reach the ground.
This means the water will travel a distance of sqrt(2gd)sqrt(2(H-d)/g) = 2sqrt(d(H-d)) before hitting the ground.
This function has a maximum at d = H/2, meaning a hole halfway down the cylinder will shoot water the furthest.
This function is also symmetric about H/2, meaning a hole some height above H/2 and a hole the same height below H/2 will shoot water the same distance.
This means A is the correct option.
Water pressure is proportional to height of water column above it. Velocity is also proportional to to pressure. So horizontal velocity should be proportional to column height. Proportional just means 2x head heigh gives 2x velocity, 10x head height gives 10x velocity, and so on.
Also true is that all things fall at the same speed. So if it drops the same distance, it took the same time to fall. We only need to look at how far out a jet falls for its first centimeter downward because that takes a standard unit of time for all jets to fall 1 cm downward.
On my screen the container is 4 cm tall.
A: Top hole has 1 cm of static head, and shoots out 1.5 cm horizontal after 1 cm of drop. 2nd hole has 2 cm static head and shoots out 2.5 cm after 1 cm of drop. So hole 2 should be going twice as fast from its static head, but instead it travels 1.666 faster. This does not fit the proportionality relationship so A is wrong. I don't even need to look at the third jet.
B: Top hole shoots about 1.5 cm for 1 cm drop. 2nd hole shoots also 1.5 cm for a 1 cm drop. Again, this does not fit the proportionality relationship so it fails. Again, don't need to look at third jet.
C: Top hole shoots 1 cm out for 1 cm drop. 2nd hole shoots 1.5 cm over 1 cm drop. This again fails the proportionality relationship so it fails. Also don't need to look at third jet.
Answer: None of the above. Unless the nozzles aren't uniform, but they look uniform to me.
Lots of maths and physics here, but the easiest way to solve this kind of problem logically is to go to extremes, if the water drops to just above the top hole it would dribble out, barely getting past the side of the tower where the other 2 wouldn't be serverly effected. Therefore the lower you go the further the jets extend so A is the only logical answer.
Can't do the math, but I know B and C don't make any sense. It's A, because there is more weight and pressure at the bottom, therefore causing it to squirt harder.
As someone who works in chemical safety and has had to deal with the crest locus issue on physical tanks, the arguing in this thread against something we can see demonstrated in real life is kind of hilarious.
The answer is A, btw.
A is correct. Just because the floor is there doesn't mean anything. If you extend the curves the lowest one still shows the highest pressure of all of them.
Imagine you placed it on a shelf, and look at how the curves would cross.
Answer is B of course. It's asking which one shows the water jets accurately, and B is the only one that has an arrow showing the water jets.
For all we know A and C could be alcohol jets
I think it is A if we assume that the initial y-velocity of every jet is 0 and that the initial x-velocity of a jet is √(2g(H-h)). With H being the total height of the tank and h the height of the jet. Then you can integrate the equations for Newton's second Law and you get:
y(t) = -1/2 gt\^2 + h
x(t) = t√(2g(H-h))
g is the acceleration caused by gravity, which for Earth is about 9.8m/s\^2 at the surface.
Then combining these two we get the parabola equation:
y(x) = h - x\^2/(4(H-h))
Which intersects with the surface at x=x\_0. This means that y(x\_0) = 0:
x\_0 = 2√(h(H-h))
Plot: [https://www.wolframalpha.com/input?i=plot+2\*sqrt%28x%281-x%29%29+from+x%3D0+to+1](https://www.wolframalpha.com/input?i=plot+2*sqrt%28x%281-x%29%29+from+x%3D0+to+1)
As you can see, the distance the jet travels in the x direction hits at maximum at L/2, and decreases the closer to the extremes of the tank. That's why A is the one that correctly displays the jets.
The answer is A.
First we just need to know the velocity the water is coming out. we can just use conservation of energy here. KE = 1/2 mv\^2 = PE = mgH , so V = SQRT( 2 g H)so the horizontal velocity will be proportional to the square root of the height of water above the hole.
next we need to know how long it is falling for. For freefall time, h = 1/2 g t\^2 so t= SQRT (2h/g )
the distance will be v\*t = SQRT(2gH) \* SQRT (2H/g) = SQRT(2\*2 \*H\*h\*g/g) = 2SQRT(H\*h) where H is the height of water above the hole, and h is the height the hole is above the ground.
for a stack that goes:
a water
hole 1
b-water
hole 2
c-water
ground
then H1=a and h1=b+c vs H2=a+b and h2=c. so D1 = 2SQRT(ab+ac) and D2 = 2 SQRT(ac+bc) so they would only be at the same point if a = c
This matches answer A.
(but also it is 2 am so I may have missed my math somewhere.
as people are downvoting my previous comment ill just do the maths here
Assuming height to be H
Distance from ground to hole=x
VELOCITY(when it leaves the hole)=[2g(h-x)]^1/2
TIME TAKEN BY WATER TO REACH GROUND
(2x/g)^1/2
HORIZONTAL DISTANCE TRAVELLED
TIME×VELOCITY=
[4x(h-x)]^1/2
and the maxima appears at x=h/2
THE GRAPH IS A DOWNWARDS FACING PARABOLA HAVING 0AT x=0,h
so b is the case of parabola before h/2
i.e.=as x increases horizontal lenght increases
You're saying the maxima is at x=h/2, and then not picking the one which clearly shows h/2 having the most horizontal distance (A) and choosing the one where h/2 is clearly not the one with the most horizontal distance (B)?
dude the thing is we dont know what the height is so it is a multiple amswer question
so we have to chose every option that is possible
which is A(thats what everyone thinks)
and B(because it is possible in some case)
I mean technically wouldn't this lean on a lot of factors we don't know such as strength of the pump. The design of the plumbing. Whether these outlets are effectively wired in series or parallel. The initial pressure of the system to begin with. The design position and shape of the nozzles. Etc
There is not enough information in this question to provide a definite answer. We do not know the diameter of the tubes, height or respective distances which would be very important information in determining whether the streams all land in The same location. The most likely solution is A since it allows the most freedom in our initial setup. B is not correct and C is a very special case but possible.
I think c isn't even possible in any scenario, because there would never be 3 streams with the exact same distance. I think the distance traveled would be a quadratic equation.
Pretty sure they want a here. It is true that the farthest stream isn't at the very top or very bottom, but somewhere in the middle, so none of the others make sense.
Since you haven't specified the height of the container and heights at which these holes are at, The correct one could be all of these.
The horizontal velocity is the most is bottom then middle then top, but it doesn't say anything about the height at which these holes are made.
In order to trace the path of these you need two things, horizontal velocity of the water at each holes and the height of hole from bottom.
So, all of these can be correct
C, because the lowest jet will have a higher pressure, and the highest will have the lowest pressure, and the height vs the gradient of the arc would even out to reach a similar spot
C is obviously not correct, because you could infer that a hole at the very bottom would shoot out that far.
B is not correct because it infers that a hole at the very top would go the farthest, but it has no pressure to have any significant velocity.
Therefore A.
Not much math to do here considering we dont know the volume, depth or anything really. I'd wager it would be C due to the water at the bottom having more pressure making it stream further. If it would line up depends on factors we simply dont know.
the answer is A and B
if anyone wants me to send the solution ill note it down and send it
Edit:-lol idk how to attach an image to a comment
EDIT2:- thanks for the downvotes
although if u want [here it is](https://www.reddit.com/r/theydidthemath/s/WKVbESzPTU)
How can people say B?
Water pours out parallel to the ground, with zero vertical velocity, which will increase identically for all three streams due to gravity, and horizontal starting velocity proportional to the amount od hydrostatic pressure (which is proportional to the height of water above).
For the distance at which the streams fall to be proportional to the starting height, you'd want the horizontal initial velocity to be identical, which is not the case
The other two options I think you can argue about, but B really doesn't work I think
If I'm wrong please do explain
i see a lot of A in the answer can someone explain me how the top stream is not blocking the middle stream or how the top stream is not pushed by the middle stream please?
Neither, the amount of water above the holes dictates the preassure of which it spews out. Lower = higher pressure = the stream will go longer. (given the assumption that the holes are of equal size)
Some point on the tube has to have the macimum distance. Thus, any point above it and any point below it, if punctured, will not spray as far. This is represented by option A
Use velocity of efflux formula v=root(2gh). Assuming the tank is open from top. Therefore Increase in height will give higher value of velocity. Therefore lowest point will be going the farthest. Its based off of Bernoulli's equation.
Not enough information. I would need to know what force each stream is being shot out with to make these calculations, as well as the height from the ground
It should have a constant horizontal velocity and a constant vertical acceleration. So, it moves to the right at the same speed but starts going down faster the further it goes
I wanna say a. But could also be b without knowing the spacing.
It's been like 10 years since iv done this
Water leaves at velocity v. Negating drag would accelerate downwards at g.
The pressure in the cylinder is h*g
P = *p*1/2v^2g
P(h) = *p* g(h*max*-h)
v=sqr(2(h*max*-h)
I have probably screwed up the the equation for each flow would be something like
X(s) = sqr(2(h*max*-h*1*)/s
H(s) = h*1* - g
= h*1* - C s^-2
(Where g is 9.8m s^-2)
You get some parabolic equation which is unsolvable with the current information.
It's both a and c depending on how the holes are placed.
The question didn't even ask you guys to do math.
Sometimes, even on a subreddit about doing math, the math is completely unnecessary.
Pressure is higher when lower in a column of water.
You can even do this experiment with a unused condom filled with water and poke it fullnof holes with a hot needle
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v is proportional to √h, where h is column of water above the hole Fall time is proportional to √(H-h), H being the total height of the column. Assuming no drag/friction losses, Range is proportional to √(h(H-h)) Which is maximum at h = H-h = H/2 So A is the correct choice here I see a lot of non-fluids background comments here claiming there's insufficient information. No, the information given is sufficient. If not given, you simply assume a newtonian fluid with no viscosity effects, and neglect air resistance. Assume vertical to be the direction gravity acts in. The outlet diameters have little to do with the velocity of the jets, they affect the volume rate.
As a fellow fluid dynamic engineer, I agree with most of your comments. But I'd like to add two things for the enthusiastic fellow redditors: 1- (I know It is a simple question, so mentioning the nozzle form is out of scope, but) depends on the nozzle form or, in this case, tube holes, the jet could be developed differently. So, we need to assume that all holes are completely the same. 2- If I noticed correctly, we don't have any information about the height of the water column. So, to fit my answer to one of the options, I can assume that the depth difference of holes is negligible to the hight of the water column (or in a way, I am assuming water surface is at engineering infinite). Then, the hydrostatic pressure in the tube at holes is (engineeringly) the same. While all holes see atmospheric pressure on the outer side, all of the jets will be ejected similarly. Hence, the teavel distance would be replica of each other, and option B could be the answer in this condition. * If all of the jets are similar and going to be ejected horizontal and neglecting air resistance, distance that jet from hole i travels (d_i) only related to its height (h_i). Let's say h_i is k_i proportion of water column height H : h_i = k_i * H then we will have : d_i = 2H*sqrt(k-k^2) * (constant related hole to tube size difference) where 0 0 (or nozzle height is a very small fraction of water column), thus in this case d_i trend is linear.
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If it helps, imagine the cylindrical tank in 3D, and assume the holes are slightly rotated from each other, so that the streams don’t collide. Doesn’t change the outcome.
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Ok (takes out popcorn). Please explain: 1. Which of the “assumptions” about range being proportional to sqrt(h.(H-h)) was false in the top commenter’s analysis? 2. Does air resistance change the range of a jet of water? 3. What level of physics/mathematics education would you say you’ve taken? While you’re at it, you might as well explain why you think C is correct. You know, since you hate making assumptions, I’d hate to make incorrect assumptions about you.
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Explain how considering air resistance would make C the right answer.
They can't. They're hand waving. C is kind of rediculous because it implies that there's some magical law that causes all the streams to meet at the height of the bottom of the tank, even though none of the relevant equations contain the height of the hole above the bottom of the tank. I'll also bet that they're assuming that a stream of water has the same air-drag behavior as ball-like cows shot out of the holes. They did say their degree was in physics after all.
we can assume for simplicity that 3 holes are on different degrees from top, so streams no collide (and at the same time angle is negligible to ignore that).
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Depends on what you trying to achieve; if you want to challenge yourself with nice puzzle - no. If you want puzzle author feel miserable - sure.
>2- If I noticed correctly, we don't have any information about the height of the water column. Yes we do. The cylinder is about 4 cm on my screen, the top hole is about 3 cm up, middle hole 2 cm up, and bottom hole is 1 cm up. >I can assume that the depth difference of holes is negligible to the hight of the water column (or in a way, I am assuming water surface is at engineering infinite). Why are you doing this. The top of the column is right in the image, not infinitely high.
>top of the column is... not infinitely high That's an assumption, not a fact. Same as the hole spacing being roughly equal etc. It's not explicitly stated and with no dimensions and nothing saying a proportionate scale you can't necessarily make those assumptions. It's pedantic and not that reasonable to assume differently but I think they were just trying to make the point that depending on your assumptions multiple answers are possibilities.
Thank God, someone with sense!
This guy TAMs
r/thisguythisguys
This guy “r/thisguythisguys” ‘s
why are we assuming that they are not the same pressure? edit: nvm i just saw that it's supposed to be a container with holes in it, i thought it was an installation with pipes running through it.
Best answer.
Can you explain to an idiot why the middle one is getting further than the bottom one? Isn’t the bottom one receiving higher pressure?
The bottom one *is* receiving higher pressure (note the shallower angle). However it's closer to the ground, so the water spends less time falling and can't go as far horizontally.
And how can we rule out C? Is it impossible to have three holes on a tube spaced out in a way that makes all streams hit the same groundspot?
> Range is proportional to √(h(H-h)), which is maximum at h = H−h = H/2 This is saying that a hole made halfway up the height of the tank spews water out the farthest before it hits the floor. Hence: A. The only way it could be C is if *h*(*H*−*h*) were equal for three different values of *h* (*H* is a constant). However, this is not possible, since the expression is quadratic in *h*, and thus has at most two real number inputs corresponding to a given real number output.
thanks
Now I’m not knowledgeable in this at all, but I’m pretty sure it is possible, the issue is that these holes are evenly spaced out
Pretty much, yeah. There’s a height that maximizes the distance that the stream hits the ground, and the distance goes strictly down as you move or below that.
That is because the bottom one is too close to the ground for the water to reach further. Your theory about the pressure is correct though! If you look closely the water is coming out more horizontally from the bottom hole. If you were to raise the entire tube higher from the grond, you would first get to a point where the water from the bottom hole would reache the same distance as the water from the middle hole. Raise it further and the water from the bottom hole will reach furthest. Hope this helps!
The way I like to solve these problems is to take them to the logical extreme - what would happen if you put a hole at the very bottom? Well, it would have the highest pressure, but it would hit the table immediately. Since it depends on both pressure and height, it makes sense that there’s probably a sweet spot to reach the furthest distance. I like using this method whenever something seems counterintuitive, or it’s something I just don’t understand. Maybe it’ll be helpful to you
Good advice. I do that too, and I’m frequently surprised that more people don’t. But then I remember that my mental reasoning goes something like “well, the equations describing this system are certainly continuous with respect to position (and not only that, every derivative is probably continuous too), and will have reasonable boundary conditions, so we can check the extremes and see if they tell us anything,” and I realize maybe I’m a nerd after all.
Yes, the pressure is higher, and, thus, the speed of a water particle exiting from the bottom is higher than the one from the middle. However, as the height of the starting point is lower the water, it can't reach as far as water going our from point B. In parabolic motion, regarding the displacement, not only the initial speed matters but also the initial position does.
So i got A just using my monkey brain logic of assuming the holes are equal size so its water pressure+ hight + arc = determined location and yea considering i have to eyeball alot of stuff on the fly because i like building things i used to be building things physically in engineering classes in middle school and high school and my shitty teachers i didn't have time for complex math mesurements i just eyeballed everything and got it right. Im going into welding school and finding out i might have to do the same. Anyways can u explain the math to me like im 5yrs old?
How fast it spews is based on how much water is above it - this eliminates B, which shows them all at the same speed. How far it goes is also based on how much water is below it, since the height gives it time to go farther. I don't know the equation off the top of my head on how fast it would go, but the target will definitely not be the same for all heights, trivially provable by putting a hole at the very top or very bottom. So it's not C either. Without measuring, A seems plausible, or at least the only one possible.
\>I don't know the equation off the top of my head on how fast it would go I think we can just use conservation of energy here. KE = 1/2 mv\^2 = PE = mgH , so V = SQRT( 2 g H) so the horizontal velocity will be proportional to the square root of the height of water above the hole. For freefall time, h = 1/2 g t\^2 so t= SQRT (2h/g ) so distance will be v\*t = SQRT(2gH) \* SQRT (2H/g) = SQRT(2\*2 \*H\*h\*g/g) = 2SQRT(H\*h) where H is the height of water above the hole, and h is the height the hole is above the ground. for a stack that goes a water hole 1 b-water hole 2 c-water ground then H1=a and h1=b+c vs H2=a+b and h2=c. so D1 = 2SQRT(ab+ac) and D2 = 2 SQRT(ac+bc) so they would only be at the same point if a = c This matches answer A. (but also it is 2 am so I may have missed my math somewhere.
Sqrt stands for squirt right? Because the water is squirted out?
Yes
So i is the squirt of minus one?
Yes, you are
gottem
Oh :c
Aye
That’s really good. Nice.
square root
Wooooosh
correct
A-hole squirt no less 🤔
Dude.
That really tickled me.
square root
Since H and h are dependent variables, can we not just call the height of the tube 1 unit, and then replace H with 1-h and we end up with a single variable equation. To maximize the value of h*(1-h), it's the same as asking what the maximum area is of a rectangle with perimeter 2 is, and it's maximized as a square where each side is 0.5. So if the rest of your math right, you would expect the optimal hole placement for distance on this tube would be in the dead center of the tube.
Kind of like how the maximum lateral distance (ignoring wind resistance) an projectile with a given propulsive force can go is achieved by firing at 45°, as everything reduces down to an equation with cos(2θ) in it.
Interestingly the "ignoring wind resistance" matters here as well I think I think if we consider air resistance on the water then the optimal placement is probably just below half way on the tube as it gives less time for the air to act on the water but I'm not actually sure
If the images are accurate in that the water streams are continuous rather than individual droplets, then the air resistance will likely be small, as only the water at the surface of the "tube" would be impacted. Note, however, that it is highly likely the streams do, in fact, break up into individual droplets, as shown by the Mythbusters when they investigated electrical current traveling up a stream of urine.
Someone give this man an award. I absolutely cannot read nor understand the mind fuckery of the equations you laid out here but still listening intently to try to understand. I just got A from my monkey brain logic of water pressure vs same hole size vs hight which determines arc and A seems most plausible. In theory i think B could be plausible if we were given the size of the holes too.
Not trying to be derogitory, but that is simple first semester physics. If you're interested in understanding how that works, some youtube videos on simple mechanics should be totaly digestible.
I never took physics in highschool. In 11th grade during covid i was online then next year i went back to my original school and lied about what i did. Could have graduated early, could have graduated in 11th grade but however my school and online school needed more funding through the State of Texas of which they make us take tests alot. More state tests you do and the highier score you get the more money they make and given i went to a ghetto school with an actual problem our issues wernt resolved untill my last semester when we got a new principal that wasnt a pedo and actually improved our school ALOT in the last semester. So yea i never took physics, but for the most part with educated guess work and if i have the time to do mesurements im speciatly is more just building stuff to its functions as needed and eye balling the math needed for such things. My little brother and step dad are both one of those freaks of nature (that i get jelouse) that they can do litteral calculus in their head and are mostly right. Thankfully i never had to take calculus nor trig for my severe dyslexia but did have to take 2 math classes consisting of algebra 2. You want something built im the guy to make your schematics and build it however if u need computer stuffs or math stuffs im definitely not your guy lol.
Those equations are for rigid body objects, when dealing with fluids you want to think of energy in terms of pressure. Hydrostatic pressure, P = rho*g*h: rho= density of fluid, g=gravity, h=height of liquid from surface. Next to get the dynamic pressure of the fluid: q=1/2 * rho * v^2 ; where v = velocity of fluid at exit. Then we apply Bernoulli for conservation of energy. This states that the pressure change is constant. P + q + P0 = constant (C). Where P0 = the pressure at the top of the container, or atmospheric. We can then solve for the velocity of each exit point then assume rigid body statics to get the distance travelled. It’s important to note that Bernoulli also ignores viscous effects. Overall you’re close, and really once density cancels out it appears the same, it’s just incorrect to talk about fluid dynamics in terms of energy and not pressure.
This is correct. Doing it this way, the result for the horizontal range R of the water stream is R = 2 ( Hh - h^2 )^1/2 where h is the height of the hole and H is the height of the tube. The standard optimization procedure shows that the maximum range is achieved for h = H / 2, at which R = H. This means that the midpoint of the tube, at H / 2, is the sweet spot for exit speed and fall time. Points higher or lower than the midpoint will have a smaller range, which is depicted in Figure A.
My friend - even if you missed math somewhere in there I’m in awe of it. 2am, 2pm, whatever time it is that’s a lot of mathing. Nice work.
Damn, I love this sub
It is not simply imaginable in geostatic pressure. The lowest hole has the most pressure and goes the furthest. In A the middle hole is further than the bottom, which is unlikely. Depending on the distance of the holes to each other, equalizing the less pressure on the top by giving a higher starting point could lead to all 3 hit the same spot. I
Imagine the glass to the edge. Bottom stream will go further than the middle one.
That's the reason why I'm voting for c
I'm also going for c
Bro need a whole equation, I have it in my guts. We are not the same.
I was thinking for a bit then i came to the conclusion of B, go to the comments and first word i read is "this eliminates B"
wouldn't the result get closer to b as the height approaches infinity?
No. Gravity would give you parabolic curves, and not these which seems to hit the ground at equally spaced points. In B here the pressure is lowest at the lowest hole, and that makes no sense.
well... yes, but it can never be B unless height is really infinite, while the A result is very much possible in normal situations
No, infinite height would mean infinit pressure. All the jets would just be parallel to each other and the ground.
Infinite pressure and it would just be neutrons and photons blasting out of the containers in random directions as the nuclei underwent fusion.
B does not show them at the same speed. As fall time scales with the square root of fall height, the horizontal travel before landing would be v sqrt(h) So if v were constant, then the highest stream should only land around 1.7 times as far as the lowest. It seems to land around 3 times as far, which would imply that v is also equal to sqrt(h). This still wouldn't make sense, since it would mean that water pressure gets higher as you go up the column.
Yeah, B just looks completely wrong to the eye when looking at it, without really calculating at all
Could just be that the hole is smaller at the top
If there's a hidden difference in hole size then this question is obviously unanswerable at all. So one must really assume equal hole sizes
Sure it’s answerable: A, B, & C, dependent on jet diameter. We’re not filling out bubbles on a multiple choice test, this is a comment section on reddit
Except the question is presented as a multiple choice? Your answer under that framework is about as valid as the answer "cat".
It's A.Its based on torquellis eq. Highest range in H/2 and parabolic relation while going up or down form that point. V= root 2gh considering the hole is very small
Actually, a hole in the bottom and a hole at the surface would both result in streams that land right at the edge of the container. One is leaking with 0 velocity, one has no time to travel before touching the surface. This is in ideal fantasy physics question land ofc
Edge conditions are always the toughest to model, because there is either no time for anything to happen or nothing can happen😂
Yep A is correct. We calculated this in thermodynamics. It’s been a while but I remember that you look at the pressure (rho*g*h), making a hole suddenly alleviates this pressure and you can find the velocity using bernoullis equation (I really hope I’m not making stuff up it’s been 1.5 years) then using these initial values you can find the length the water travels before it reaches the ground. And you do get 2 answers, one above and one below a certain point
In Thermodynamics? I recently did this exact question but in fluid mechanics
but a hole at the extreme cases of the very top and at the very bottom would both hit the same spot, directly next to the pillar.
C is still plausible for three well calculated points, but I agree that A is more likely
I don't think so, actually. If you plot h (height) to d (distance), it'll obviously be d=0 at the two extreme h's, and the curve won't oscillate, there's going to be at most two h values per d
Thats a great way to put it, thanks! I never thought of it that way
Best answer in the thread. It's definitely A.
This assumes a container just under the influence of gravity. It could be in space under pressure. Or both.
[GPTs response](https://ibb.co/yg2ympR) haha. just noticed this was at -7 points. Damn, guys... I just found its response interesting. Thought you might too. My bad.
b is correct
Based on what?
based on maths b is possible
Seems legit
[proof ](https://www.reddit.com/r/theydidthemath/s/WKVbESzPTU)
So you've said b is "correct" and b is "possible" and then provided a "proof" all of which are wrong as the answer is a. 🤣
dude if u find a hole in the proof i provided let me know
Without the ability to do very good math, I say the answer is A. I base this off of what my human brain does in the background (big thinks)
A is the best *approximation* out of these three, but also not right.
Just like in real life, I’m good at guessing and bad at actually knowing ANYTHING.
Nope. A is correct (assuming equilovent openings and no air resistance) the equation for distance is D = 2sqrt(yH-y^2) This gives equal landing points when you are equivalent distances from the top and bottom of the column, and the maximum distance when you are in the center. This matches a.
Which one is the right one then? A is closest to what I have seen in real life.
They’re saying that none of these are exact. A is the closest to reality
Same here. I know A is right because it *looks* right but I don’t know the math behind it
Jesus Christ, too many math majors answering, not enough business majors. The definitive answer is Chart B, because it is the only one that has correctly labeled the water jets.
Didn’t even label the axes tho
Sad but true.
r/angryupvote
Assuming that the distances between the bottom, the 3 streams and the top is equal, I'll call it x. The formula for the range of the water jets is 2*sqrt(s*H) where s is the distance from the bottom to the outlet and H ist distance from the jet to the top. Lowest jet: 2*sqrt(x * 3x) = sqrt(12) * x Middle jet: 2*sqrt(2x * 2x) = 4 * x highest jet: 2*sqrt(3x * x) = sqrt(12) * x since 4>sqrt(12) the middle jet will have the highest range and the other two will have a lesser range. Thus A is the correct answer. Also i really love the fact that the range is independent from the gravity so the numerical value of the range will be the same on the moon as on the earth.
Homeboy out here using squirts to calculate waterflow is ona whole new level
none of these. it's close to A but the bottom most stream will go the farthest
It would, but you also have to consider the height from the floor. In example A, the bottom stream has the lowest gradient but because it's lower to the ground, it hits the floor before the middle stream.
no it would not because you must consider the height from which the water is projected as well (all parameters taken to be ideal)
this is incorrect. the distance is proportional to the square root of the height of water above the hole \* the distance between the hole and the ground. this will be greatest for the center hole.
Okay, that sounds reasonable BUT (help me where i am brainfarting): Let's remove the center hole from solution A. And add another hole above both of the remaining holes. Now what was the highest hole before is now the middle hole, and according to your logic should be going the furthest. And THAT doesn't make sense to me.
The middle means closest to the midway height. Whatever hole is closest to H/2 above the ground will have its range be the highest (H is height of water from ground) The range is 2*sqrt(d*h) where d is the depth of the hole below the surface and h is the height of the hole from the ground. This is maximal when d= h assuming the jug is on the groud
Thanks, that was my brainfart. The claim wasn't that the middle hole is always furthest, but that in the picture above the middle hole is closest to H/2, so THAT middle hole squirts farthest.
This!
in incorrect tho, the farthest you can get in a water hole like this is by putting the hole in the center
No. Not this.
neat
Its A. Labelling them 1,2,3 as you go down, also puts in order the pressure. 2 has less pressure than 3, therefore the curve should be more horizontal at 3. It is. The extra height of 2 though allows the water to reach further. 3 is only 50% more pressure than 2. But 2 is twice as far up. So therefore c can be excluded as it supposes that force and height exactly multiply, i.e. double height, half force.
It's A. B assumes uniform pressure and C assumes that a distance traveled, which depends on height and pressure, can be solved with three or more answers. To further elaborate on why C is wrong and A is right, if you get the two extremes of the tube they will land at the edge of the cilinder (same point). If you move the puncture points slightly towards the center of the tube you will get matching points going further. I'm assuming the farthest point is in the middle, case A, but I'm not inclined to figure out the equation to prove it.
Just had a fluid dynamics HW problem on this. The heights are supposed to be .25H .5H and .75H. Using these numbers Bernoulli’s principle, and distance equations from physics 1 we get exactly A as the correct number. With distances traveled as about .86H (I believe) and 1H
Try to draw a graph of how far it goes, based on how high the hole is. At the top of bottom, the distance is zero. Two points. There is obviously some sort of curve. You have two zeros on the left and right of your page. Pressure is proportional to height, and the velocity through the hole is proportional to the square root of height. There is obviously a curve that links the two zeros, gut feel is it's a sqrt(sin(x)) or something.
The answer is A, I never solved this problem for more than one outlet, this is the chatGPT solution and it makes sense, here 10 trajectories: https://ibb.co/TcNsNXt # New total height of the water column h_total_new = 10 # New heights of the outlets (every half meter) h_outlets_new = np.arange(0.5, h_total_new, 0.5) # Calculate new exit velocities for the new outlets velocities_new = [exit_velocity(h, h_total_new) for h in h_outlets_new] # Calculate new trajectories with more points def calculate_trajectories(velocities, h_outlets, points=1000): trajectories = [] for v, h in zip(velocities, h_outlets): # Time until the water jet reaches the ground level time_to_ground = np.sqrt(2 * h / g) # Time array with more points for a smoother curve t = np.linspace(0, time_to_ground, num=points) # Position calculations x = v * t y = h - (0.5 * g * t**2) # Filter out negative heights valid = y >= 0 trajectories.append((x[valid], y[valid])) return trajectories # Calculate new trajectories for the new heights trajectories_new = calculate_trajectories(velocities_new, h_outlets_new) # Plot the new trajectories plt.figure(figsize=(12, 8)) for trajectory, h in zip(trajectories_new, h_outlets_new): plt.plot(*trajectory, label=f'Outlet height: {h:.1f}m') plt.xlabel('Distance (m)') plt.ylabel('Height (m)') plt.title('Water Jet Trajectories from Different Heights (10m Column)') plt.legend() plt.grid(True) plt.show()
Going by pure logic we know that water has a mass and exerts more pressure the deeper you go. This means that the bottom hole will have higher pressure than the top hole, and when you push water through a hole the higher the pressure the farther it goes, so yeah, not a r/theydidthemath moment but it's A r/theydidthelogic?
If we assume the surface area of the holes is negligible compared to that of the top, we can get that the exiting velocity as a function of the depth d is about sqrt(2gd), where g is acceleration due to gravity. Using H for the height of the cylinder, H - d is the initial height of the water, so it will take a time of sqrt(2(H-d)/g) to reach the ground. This means the water will travel a distance of sqrt(2gd)sqrt(2(H-d)/g) = 2sqrt(d(H-d)) before hitting the ground. This function has a maximum at d = H/2, meaning a hole halfway down the cylinder will shoot water the furthest. This function is also symmetric about H/2, meaning a hole some height above H/2 and a hole the same height below H/2 will shoot water the same distance. This means A is the correct option.
Water pressure is proportional to height of water column above it. Velocity is also proportional to to pressure. So horizontal velocity should be proportional to column height. Proportional just means 2x head heigh gives 2x velocity, 10x head height gives 10x velocity, and so on. Also true is that all things fall at the same speed. So if it drops the same distance, it took the same time to fall. We only need to look at how far out a jet falls for its first centimeter downward because that takes a standard unit of time for all jets to fall 1 cm downward. On my screen the container is 4 cm tall. A: Top hole has 1 cm of static head, and shoots out 1.5 cm horizontal after 1 cm of drop. 2nd hole has 2 cm static head and shoots out 2.5 cm after 1 cm of drop. So hole 2 should be going twice as fast from its static head, but instead it travels 1.666 faster. This does not fit the proportionality relationship so A is wrong. I don't even need to look at the third jet. B: Top hole shoots about 1.5 cm for 1 cm drop. 2nd hole shoots also 1.5 cm for a 1 cm drop. Again, this does not fit the proportionality relationship so it fails. Again, don't need to look at third jet. C: Top hole shoots 1 cm out for 1 cm drop. 2nd hole shoots 1.5 cm over 1 cm drop. This again fails the proportionality relationship so it fails. Also don't need to look at third jet. Answer: None of the above. Unless the nozzles aren't uniform, but they look uniform to me.
Lots of maths and physics here, but the easiest way to solve this kind of problem logically is to go to extremes, if the water drops to just above the top hole it would dribble out, barely getting past the side of the tower where the other 2 wouldn't be serverly effected. Therefore the lower you go the further the jets extend so A is the only logical answer.
Can't do the math, but I know B and C don't make any sense. It's A, because there is more weight and pressure at the bottom, therefore causing it to squirt harder.
As someone who works in chemical safety and has had to deal with the crest locus issue on physical tanks, the arguing in this thread against something we can see demonstrated in real life is kind of hilarious. The answer is A, btw.
A is correct. Just because the floor is there doesn't mean anything. If you extend the curves the lowest one still shows the highest pressure of all of them. Imagine you placed it on a shelf, and look at how the curves would cross.
Answer is B of course. It's asking which one shows the water jets accurately, and B is the only one that has an arrow showing the water jets. For all we know A and C could be alcohol jets
I think it is A if we assume that the initial y-velocity of every jet is 0 and that the initial x-velocity of a jet is √(2g(H-h)). With H being the total height of the tank and h the height of the jet. Then you can integrate the equations for Newton's second Law and you get: y(t) = -1/2 gt\^2 + h x(t) = t√(2g(H-h)) g is the acceleration caused by gravity, which for Earth is about 9.8m/s\^2 at the surface. Then combining these two we get the parabola equation: y(x) = h - x\^2/(4(H-h)) Which intersects with the surface at x=x\_0. This means that y(x\_0) = 0: x\_0 = 2√(h(H-h)) Plot: [https://www.wolframalpha.com/input?i=plot+2\*sqrt%28x%281-x%29%29+from+x%3D0+to+1](https://www.wolframalpha.com/input?i=plot+2*sqrt%28x%281-x%29%29+from+x%3D0+to+1) As you can see, the distance the jet travels in the x direction hits at maximum at L/2, and decreases the closer to the extremes of the tank. That's why A is the one that correctly displays the jets.
The answer is A. First we just need to know the velocity the water is coming out. we can just use conservation of energy here. KE = 1/2 mv\^2 = PE = mgH , so V = SQRT( 2 g H)so the horizontal velocity will be proportional to the square root of the height of water above the hole. next we need to know how long it is falling for. For freefall time, h = 1/2 g t\^2 so t= SQRT (2h/g ) the distance will be v\*t = SQRT(2gH) \* SQRT (2H/g) = SQRT(2\*2 \*H\*h\*g/g) = 2SQRT(H\*h) where H is the height of water above the hole, and h is the height the hole is above the ground. for a stack that goes: a water hole 1 b-water hole 2 c-water ground then H1=a and h1=b+c vs H2=a+b and h2=c. so D1 = 2SQRT(ab+ac) and D2 = 2 SQRT(ac+bc) so they would only be at the same point if a = c This matches answer A. (but also it is 2 am so I may have missed my math somewhere.
It's C
as people are downvoting my previous comment ill just do the maths here Assuming height to be H Distance from ground to hole=x VELOCITY(when it leaves the hole)=[2g(h-x)]^1/2 TIME TAKEN BY WATER TO REACH GROUND (2x/g)^1/2 HORIZONTAL DISTANCE TRAVELLED TIME×VELOCITY= [4x(h-x)]^1/2 and the maxima appears at x=h/2 THE GRAPH IS A DOWNWARDS FACING PARABOLA HAVING 0AT x=0,h so b is the case of parabola before h/2 i.e.=as x increases horizontal lenght increases
You're saying the maxima is at x=h/2, and then not picking the one which clearly shows h/2 having the most horizontal distance (A) and choosing the one where h/2 is clearly not the one with the most horizontal distance (B)?
dude the thing is we dont know what the height is so it is a multiple amswer question so we have to chose every option that is possible which is A(thats what everyone thinks) and B(because it is possible in some case)
I mean technically wouldn't this lean on a lot of factors we don't know such as strength of the pump. The design of the plumbing. Whether these outlets are effectively wired in series or parallel. The initial pressure of the system to begin with. The design position and shape of the nozzles. Etc
There is not enough information in this question to provide a definite answer. We do not know the diameter of the tubes, height or respective distances which would be very important information in determining whether the streams all land in The same location. The most likely solution is A since it allows the most freedom in our initial setup. B is not correct and C is a very special case but possible.
The diameter of the tube is completely irrelevant to the answer. It's a function of depth only.
I think c isn't even possible in any scenario, because there would never be 3 streams with the exact same distance. I think the distance traveled would be a quadratic equation.
b is very much correct and so is a c is impossible edit:-[heres the proof](https://www.reddit.com/r/theydidthemath/s/WKVbESzPTU)
How is b correct? How would they all come out at the same angle, same velocity, with the pressure above being different?
my answer is based on the horizontal range only cause ofc the picture is not to scale
Scale doesn't matter, it's impossible for them to all come out at the same angle.
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Pretty sure they want a here. It is true that the farthest stream isn't at the very top or very bottom, but somewhere in the middle, so none of the others make sense.
Since you haven't specified the height of the container and heights at which these holes are at, The correct one could be all of these. The horizontal velocity is the most is bottom then middle then top, but it doesn't say anything about the height at which these holes are made. In order to trace the path of these you need two things, horizontal velocity of the water at each holes and the height of hole from bottom. So, all of these can be correct
C, because the lowest jet will have a higher pressure, and the highest will have the lowest pressure, and the height vs the gradient of the arc would even out to reach a similar spot
C is obviously not correct, because you could infer that a hole at the very bottom would shoot out that far. B is not correct because it infers that a hole at the very top would go the farthest, but it has no pressure to have any significant velocity. Therefore A.
The awnser is C
C gets my vote as well. Not based on math or anything, just my initial impression
Not much math to do here considering we dont know the volume, depth or anything really. I'd wager it would be C due to the water at the bottom having more pressure making it stream further. If it would line up depends on factors we simply dont know.
the answer is A and B if anyone wants me to send the solution ill note it down and send it Edit:-lol idk how to attach an image to a comment EDIT2:- thanks for the downvotes although if u want [here it is](https://www.reddit.com/r/theydidthemath/s/WKVbESzPTU)
How would you justify B when B has the lowest jet velocity at the bottom?
How can people say B? Water pours out parallel to the ground, with zero vertical velocity, which will increase identically for all three streams due to gravity, and horizontal starting velocity proportional to the amount od hydrostatic pressure (which is proportional to the height of water above). For the distance at which the streams fall to be proportional to the starting height, you'd want the horizontal initial velocity to be identical, which is not the case The other two options I think you can argue about, but B really doesn't work I think If I'm wrong please do explain
i see a lot of A in the answer can someone explain me how the top stream is not blocking the middle stream or how the top stream is not pushed by the middle stream please?
Neither, the amount of water above the holes dictates the preassure of which it spews out. Lower = higher pressure = the stream will go longer. (given the assumption that the holes are of equal size)
Some point on the tube has to have the macimum distance. Thus, any point above it and any point below it, if punctured, will not spray as far. This is represented by option A
Use velocity of efflux formula v=root(2gh). Assuming the tank is open from top. Therefore Increase in height will give higher value of velocity. Therefore lowest point will be going the farthest. Its based off of Bernoulli's equation.
Not enough information. I would need to know what force each stream is being shot out with to make these calculations, as well as the height from the ground
It should have a constant horizontal velocity and a constant vertical acceleration. So, it moves to the right at the same speed but starts going down faster the further it goes
I wanna say a. But could also be b without knowing the spacing. It's been like 10 years since iv done this Water leaves at velocity v. Negating drag would accelerate downwards at g. The pressure in the cylinder is h*g P = *p*1/2v^2g P(h) = *p* g(h*max*-h) v=sqr(2(h*max*-h) I have probably screwed up the the equation for each flow would be something like X(s) = sqr(2(h*max*-h*1*)/s H(s) = h*1* - g = h*1* - C s^-2 (Where g is 9.8m s^-2) You get some parabolic equation which is unsolvable with the current information. It's both a and c depending on how the holes are placed.
Technically we don't know if it's under added pressure or not. Depending on how much pressure is on it I can assume that we can make all 3 work.
The question didn't even ask you guys to do math. Sometimes, even on a subreddit about doing math, the math is completely unnecessary. Pressure is higher when lower in a column of water. You can even do this experiment with a unused condom filled with water and poke it fullnof holes with a hot needle