- #1

CAF123

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## Homework Statement

A body of some mass ##m## is thrown up from the ground with a speed of ##v_o##. Suppose that upon returning to ground level, the body enters into a deep well. What is the relation between the height of the well and the time it takes until the body hits the ground?

## Homework Equations

##\ddot{x} = -g##

## The Attempt at a Solution

I am a little confused with the wording of the question. Does that mean the height as a function of time or time function of height?

I assumed the former: On entrance to the well, by symmetry, ##\dot{x}(0) = -v_o, x(0) = 0##. I have two scenarios which I think justify why ##\dot{x}(t_G) = 0##. 1)If bottom of well soft, body will just stick. 2)If bottom of well hard, body will rebound to some height but this means it's velocity has to change direction so at some instantaneous moment (which I take just before it rebounds) it's velocity is zero. So ##t_G## = time to ground = ##v_o/g##.

Let h be the height of the well. Then ##-h = -v_ot_G - \frac{1}{2}gt_G^2##. Subbing in ##t_G## gives ##h = \frac{3v_o^2}{2g}##, which doesn't make sense since the sign is wrong. I realize what I have here is an implicit function of time. Is this what the question wanted or is there something else?