# A cylinder full of water with two holes at different heights

• ValeForce46
In summary, the conversation discusses the use of Bernoulli's principle and equations of parabolic motion to solve for two unknowns, height and distance. The simplified expressions for both variables are given in terms of the given values, and a final result is confirmed to be correct.
ValeForce46
Homework Statement
To the sides of a container full of water there are two holes with a section negligible compared to the section of the base respectively at a distance of ##h_1=20 cm## and ##h_2=80cm## from the free surface of the water. Assuming that the tap pumps water in the container keeping constant the level of the liquid while water comes out from the holes and knowing that the two jet of water touch the ground in the same point (see picture), determine:
a) The height ##h## of the container;
b)the distance ##d## from the wall of the cylinder in which the holes are drilled and the point of the ground touched by the jets.
Relevant Equations
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This is what I did:
I know that ##v_1=\sqrt{2gh_1}## and ##v_2=\sqrt{2gh_2}## thanks to Bernoulli's principle.
Using equations of parabolic motion I get
##h-h_1-\frac{1}{2}g(\frac{d}{v_1})^2=0## and ##h-h_1-\frac{1}{2}g(\frac{d}{v_2})^2=0##. This means I have two equations in two unknowns ##h## and ##d##.
That's it? I feel like I'm totally wrong but I don't see other ways... (I checked dimension of both results)
the results are
$$d=\sqrt\frac{2(h_2-h_1)}{g(1/v_1^2-1/v_2^2)} = 0.8 m$$
and the height ##h=1.0 m##
Just let me know if you think I'm right :)

Looks OK to me.

Looks OK to me, too.

It's interesting to see how the expression ##d=\sqrt\frac{2(h_2-h_1)}{g(1/v_1^2-1/v_2^2)} ## simplifies if you make the substitutions ##v_1=\sqrt{2gh_1}## and ##v_2=\sqrt{2gh_2}##.

Likewise, ##h## has a simple expression in terms of ##h_1## and ##h_2##.

ValeForce46, Delta2 and mfb
TSny said:
Looks OK to me, too.

It's interesting to see how the expression ##d=\sqrt\frac{2(h_2-h_1)}{g(1/v_1^2-1/v_2^2)} ## simplifies if you make the substitutions ##v_1=\sqrt{2gh_1}## and ##v_2=\sqrt{2gh_2}##.

Likewise, ##h## has a simple expression in terms of ##h_1## and ##h_2##.
You’re definitely right. At the end I get
##d=...=\sqrt{4h_1h_2}=0.8m##
Thank you!

## 1. How does the water level change in a cylinder with two holes at different heights?

The water level in a cylinder with two holes at different heights will remain constant as long as the holes are at the same height. However, if one hole is higher than the other, the water level will decrease in the higher hole and increase in the lower hole until the levels are equal.

## 2. Why does the water level change in a cylinder with two holes at different heights?

The water level changes in a cylinder with two holes at different heights because of the principle of communicating vessels. This means that the water level in the two holes will try to equalize, creating a constant flow of water between them.

## 3. How does the size of the holes affect the water level in a cylinder?

The size of the holes does not affect the water level in a cylinder with two holes at different heights. As long as the holes are the same size, the water level will remain constant. However, if one hole is larger than the other, the water will flow more quickly through the larger hole.

## 4. What happens if I add more water to the cylinder?

If you add more water to the cylinder, the water level will increase in both holes. However, the higher hole will still have a lower water level than the lower hole due to the principle of communicating vessels.

## 5. How does air pressure affect the flow of water in a cylinder with two holes at different heights?

Air pressure does not have a significant effect on the flow of water in a cylinder with two holes at different heights. However, if one hole is significantly higher than the other, it may create a siphon effect, causing the water to flow faster through the lower hole.

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