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.