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(C) For a stone dropped from height $h$,the equation of motion is: $h = \frac{1}{2} g t^2$ ... $(i)$For a stone thrown upwards with velocity $u$,the d

Vedclass · Accepted Answer

$t = \sqrt{t_1 t_2}$

Vedclass · Answer

(C) For a stone dropped from height $h$,the equation of motion is: $h = \frac{1}{2} g t^2$ ... $(i)$For a stone thrown upwards with velocity $u$,the displacement is $-h$ (taking downward as positive): $-h = -u t_1 + \frac{1}{2} g t_1^2$,which simplifies to $h = u t_1 - \frac{1}{2} g t_1^2$ ... (ii)For a stone thrown downwards with velocity $u$: $h = u t_2 + \frac{1}{2} g t_2^2$ ... (iii)Equating $(i)$ with (ii) and (iii):$u t_1 - \frac{1}{2} g t_1^2 = \frac{1}{2} g t^2 \implies u t_1 = \frac{1}{2} g (t^2 + t_1^2)$ ... (iv)$u t_2 + \frac{1}{2} g t_2^2 = \frac{1}{2} g t^2 \implies u t_2 = \frac{1}{2} g (t^2 - t_2^2)$ ... $(v)$Dividing (iv) by $(v)$:$\frac{t_1}{t_2} = \frac{t^2 + t_1^2}{t^2 - t_2^2}$$t_1 t^2 - t_1 t_2^2 = t_2 t^2 + t_1^2 t_2$$t^2 (t_1 - t_2) = t_1 t_2 (t_1 - t_2)$Since $t_1 
eq t_2$,we get $t^2 = t_1 t_2$,so $t = \sqrt{t_1 t_2}$.

$A$ stone is dropped from a building of height $h$ and it reaches the ground after $t$ seconds. From the same building,if two stones are thrown (one upwards and one downwards) with the same velocity $u$ and they reach the earth's surface after $t_1$ and $t_2$ seconds respectively,then:

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