# Homework Help: Inverted garbage can

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1. May 30, 2017

### Buffu

1. The problem statement, all variables and given/known data
An inverted garbage can of weight $W$ is suspended in air by water from a geyser. The water shoots up the ground with speed $v_0$, at a constant rate $dm/dt$. The problem is to find the maximum height at which garbage can rides.

2. Relevant equations

3. The attempt at a solution

Suppose the garbage can is at its highest position $h$.

Then the velocity of elementary mass $\Delta m$ just before collision with the can would be $\sqrt{v_0^2 - 2gh}$ and velocity after collision would be zero.

So change in momentum would be $\Delta P = - \Delta m \sqrt{v_0^2 - 2gh}$

Or the force on the can would be $F = \sqrt{v_0^2 - 2gh} \dfrac{dm}{dt}$

Since the forces on the can is balanced, therefore $W = \sqrt{v_0^2 - 2gh} \dfrac{dm}{dt}$,

Solving for $h$ I got $\displaystyle h = \dfrac{1}{2g}\left(v_0^2 - \left(W \over \dfrac{dm}{dt}\right)^2 \right)$

This is incorrect.

Where am I incorrect ?

2. May 31, 2017

### haruspex

It all looks right to me. Do you know what the answer is supposed to be?

3. May 31, 2017

### Nidum

So where does the water go ?

4. May 31, 2017

### haruspex

It falls. I do not see any basis for claiming that the water bounces off the bucket, or in any other way makes a greater contribution to supporting it. The bucket could be significantly wider than the jet.

5. May 31, 2017

### Buffu

A clue was given; if $v_0 = 20m/s$, $W = 10 kg$, $dm/dt = 0.5 kg/s$ then $h = 17 m$.

These values does not match when I put it into my formula :(.

@Nidum It falls down under gravity.

6. May 31, 2017

### haruspex

I assume you mean W=10N.
That answer is clearly wrong. 20m/s is the speed at which the water needs to be moving when it reaches the bucket.
If you use g=9.8m/s2 and v0=27m/s it works out quite accurately to 17m.

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