If $d < d_{L}$ obtain an expression (in terms of $d, t_{1}$ and $d_{L})$ for the time $t_{2}$ the ball takes to come back to the position from which it was released.

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Solution

In elastic collision with the surface, direction of velocity is reversed but its magnitude remains the same.
Therefore. time of fall=time of rise.
or time of fall $= \frac{t_{1}}{t_{2}}$
Hence, velocity of the ball just before it collides with liquid is
$v = g \frac{t_{1}}{2}$ (i)
Retardation inside the liquid
$a = \frac{u p t h r u s t - w e i g h t}{m a s s}$
$= \frac{V d_{L} g - V d g}{V d}$
$= (\frac{d_{L} - d}{d}) g$ (ii)
Time taken to come to rest under this retardation will be
$t = \frac{v}{a}$
$= \frac{g t_{1}}{2 a}$
$= \frac{g t_{1}}{2 (\frac{d_{L} - d}{d}) g}$
$= \frac{d t_{1}}{2 (d_{L} - d)}$
Same will be the time to come back on the liquid surface.
Therefore,
$t_{2} =$ time the ball takes to came back to the position from where it was released
$= t_{1} + 2 t$
$= t_{1} + \frac{d t_{1}}{d_{L} - d}$
$= t_{1} [1 + \frac{d}{d_{L} - d}]$
or $t_{2} = \frac{t_{1} d_{L}}{d_{L} - d}$

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If d<dL obtain an expression (in terms of d,t1 and dL) for the time t2 the ball takes to come back to the position from which it was released.

If $d < d_{L}$ obtain an expression (in terms of $d, t_{1}$ and $d_{L})$ for the time $t_{2}$ the ball takes to come back to the position from which it was released.