Find the Maximum Height of an Inverted Garbage Can Suspended by a Geyser

[Buffu]

Homework Statement

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. Homework Equations

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 ?

 

[haruspex]

This is incorrect.

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

 

[Nidum]

velocity after collision would be zero.

So where does the water go ?

 

[haruspex]

So where does the water go ?

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.

 

[Buffu]

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

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.

 

[haruspex]

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.

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.