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A ball is projected from the bottom of a tower and is found to go above the tower and is caught by the thrower at the bottom of the tower after a time interval $t_{1}$ . An observer at the top of the tower sees the same ball go up above him and then come back at this level in a time interval $t_{2} .$ The height of the tower is

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$\frac{1}{2} g t_{1} t_{2}$

$\frac{g t_{1} t_{2}}{8}$

$\frac{g}{8} (t_{1}^{2} - t_{2}^{2})$

$\frac{g}{2} {(t_{1} - t_{2})}^{2}$

answer is C.

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Detailed Solution

If the ball reaches the tower with a speed of 'u'.

$t_{2} = \frac{2 u}{g}$

During descent, the motion of the ball after crossing the tower level will be equivalent to throwing a ball down with a initial velocity of 'u' downwards.

$h = u t + \frac{1}{2} a t^{2}$

$= \frac{g t_{2}}{2} [\frac{t_{1} - t_{2}}{2}] + \frac{1}{2} g {[\frac{t_{1} - t_{2}}{2}]}^{2}$

$= \frac{[t_{1} - t_{2}]}{2} . \frac{g}{2} [t_{2} + \frac{t_{1} - t_{2}}{2}]$

$= \frac{g [t_{1} - t_{2}]}{4} . \frac{t_{1} + t_{2}}{2}$

$= \frac{g [t_{1}^{2} - t_{2}^{2}]}{8}$

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