How to steal the water from a high tank

Classical Mechanics Level 2

You are trying to steal some water from a water tower. Your friend is going to climb the tower and drill a hole in the side (the bottom is too tough).

Your job is to place a bucket at the initial point of impact of the water. How far (in meters) from the side of the tower should you place your bucket?

Assumptions and Details:

  • The cylinder is completely filled with water.

  • The hole is drilled at the very bottom of the cylinder's side.

  • The hole is 1 cm 1 \text{ cm} in diameter.

  • The density of water is 1000 kg ⋅ m − 3 1000 \text{ kg} \cdot \text{m}^{-3} .

  • Atmospheric pressure is 1 0 5 Pa 10^5 \text{ Pa} .

  • The acceleration due to gravity is 10 m ⋅ s − 2 10 \text{ m} \cdot \text{s}^{-2} .

  • Air resistance can be ignored.

  • The water cylinder is not air tight at the top.


The answer is 20.

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Andrew Normand by Andrew Normand , 31, United Kingdom

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2 solutions

Rohit Gupta
Feb 3, 2017

According to the Torricelli equation, the speed of the water coming out of the small hole is 2 g h \sqrt{2gh} , here h h is the height of the water above the hole.

Let the height above the hole is h 1 h_1 and the height of hole from the ground is h 2 h_2 .

Water will come out of the hole at speed v = 2 g h 1 . v=\sqrt{2gh_1}. Now, we will use the concept of Projectile motion to calculate the distance where the bucket must to kept to collect the water.
We know that, horizontal range R R of a projectile thrown from a tower of height h h at speed u u is R = u 2 h g . R = u \sqrt{\frac{2h}{g}}. Therefore, R = 2 g h 1 2 h 2 g = 2 h 1 h 2 . R = \sqrt{2gh_1} \sqrt{\frac{2h_2}{g}} = 2 \sqrt{h_1h_2}.

Note: We can see that the distance where the water hits is double of the geometric mean of the height of water above the hole and the height of hole of the ground.

Putting the values of h 1 = 5 m h_1 = 5 \text{ m} and h 2 = 20 m h_2 = 20 \text{ m} , we get R = 20 m . R = 20 \text{ m}.

Hence, a bucket has to be placed at a distance of 20 m \boxed{20 }\text { m} from the tower.

5 Helpful 0 Interesting 0 Brilliant 0 Confused

Does the result depend on the shape of the water tank? That is, is this result true only for cylinders, or for any arbitrary shape?

Pranshu Gaba - 4 years, 4 months ago

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It works for all shapes but the hole should be very small relative to the size of the container so that you can approximate the kinetic energy of a water molecule at the top of the container to be 0.

Anupam Nayak - 4 years, 4 months ago

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I agree with Anupam. The speed with which water come out of the hole is 2 g h \sqrt{2gh} only if the size of the hole is much smaller than the size of the tank.

Rohit Gupta - 4 years, 4 months ago

Cool! I found it counter-intuitive at first, but this explanation makes sense. If the size of the hole is very small compared to the size of the container, then it doesn't matter what the shape of the container is.

Pranshu Gaba - 4 years, 3 months ago
Andrew Normand
Jan 20, 2017

The key to the problem is to realise that each point in a container of fluid is at equal potential - it takes no energy to take a particular bit of water from the top of the cylinder to the bottom, or vice versa.

This means water coming out of a hole in the side of the tower at the bottom will have the same kinetic energy, and so be moving at the same speed ,as water that has fallen from the top of the tower - except that the motion will be sideways. You can use the equation of motion v 2 = u 2 + 2 a s v^2 = u^2 + 2as to find this velocity. (If you prefer you could also solve kinetic and gravitational energy equations simultaneously.) Initial velociy u = 0, a =10N/kg and s is 5, giving a sideways velocity of 10 m/s.

Next you just need to find the time the water will take to fall to the ground, which is independent of the sideways speed. You can use the equation of motion s = u t + 1 2 a t 2 s = ut + \frac{1}{2}at^2 to find the time to fall 20 metres (again, u is 0, a is 10, this time s is 20 metres). You should find that the water takes 2 seconds to fall.

Finally, multiply the horizontal speed by the time to fall to find the distance, and hence where you should place your bucket.

3 Helpful 0 Interesting 0 Brilliant 0 Confused

It's really cool that the final answer ends up being twice the geometric mean of the only two relevant distances in the problem.

Steven Chase - 4 years, 4 months ago

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That is cool - I hadn't noticed that. I also hadn't realised that the answer is independent of the gravitational field strength.

Andrew Normand - 4 years, 4 months ago

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