- #1

- 535

- 72

## 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. What assumption must be fulfilled for the maximum height to be reached.

Ans. clue. If ##v_0=20\, m/s##, W=10kg, dm/dt=0.5 kg/s, then ##h_{max} \approx 17 \, m##

## Homework Equations

## The Attempt at a Solution

The forces acting on the garbage can are its own weight downward, and the force of water upward.

I have no proof that the force is strictly upward, but it seems natural to me, so I will make this assumption.

The force of water on the garbage can is the opposite of the force of garbage can on water. This force is determined by taking the time derivative of the momentum difference of water before and after collision:

## F_{g\rightarrow w} = (\frac{dP}{dt})_f - (\frac{dP}{dt})_i ##

To be consistent with the assumption that the water push on the garbage can is upward, the rebound direction of the geyser must go downward, otherwise it would change the direction of the force.

I get that,

##\triangle P_f = - (\triangle m) v_f##

##\triangle P_i = (\triangle m) v_i##

So the force of water on garbage can is :

## F_{w\rightarrow g} = - F_{g\rightarrow w}= (v_i + v_f) \frac{dm}{dt}##

But I don't know ##v_i## nor ##v_f## and don't see any way of knowing it, so I assume they are identical so that ## v_w = v_i = v_f ##.

So the motion of the garbage can be described by :

## \frac{W}{g} \frac{dv_{g}}{dt} = - W + 2v_w \frac{dm}{dt} ##

If I integrate through time:

## \frac{W}{g} v_g = -W t + 2 y_w \frac{dm}{dt} ##

And the garbage can reaches maximum height when ##v_g = 0 ##.

I don't think this will lead me anywhere, because I don't have any expression of water height vs time.

Could you please give me a hint on how to find maximum height ? Thanks