A boy throws a ball with speed `u` in a well of depth `14 m` as shown in (Fig. 5.191). On bounce with the bottom of the well, the speed of the ball gets halved. What should be the minimum value of `u( "in" ms^-1)` such that the ball may be able to reach his hands again ? It is given that his hands are at `1 m` height from top of the well while throwing and catching.
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Answer

`30 m s^-2`

Explanation

Let `v` is the same with which ball collides withb bottom, then
`v^2 = u^2 + 2g xx 15` …(i)
Now ball rebounds with speed `v//2`. So
`(v//2)^2/(2g) = 15 rarr ` to reach the hands of boy
`rArr v^2 = 8g xx 15` …(ii)
From Eqs. (i) and (ii), `u = 30 ms^-1`.