A stone dropped from the top of a building of height $h$ . It reaches the ground after $t s$ . From the same building if two stones are thrown (one upwards and other downwards) with the same velocity $u$ , and they reach the earth surface after $t_{1}$ and $t_{2}$ seconds respectively, then

$t = t_{1} - t_{2}$

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$t = \frac{t_{1} + t_{2}}{2}$

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$t = \sqrt{t_{1} t_{2}}$

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$t = t_{1}^{2} t_{2}^{2}$

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Solution

The correct option is C
$t = \sqrt{t_{1} t_{2}}$

If a stone is dropped from a height $h$ then

$h = \frac{1}{2} g t^{2}$ .....(i)

If a stone is thrown upward with velocity $u$ then

$h = - u t_{1} + \frac{1}{2} g t_{1}^{2}$ ........(ii)

If a stone is thrown downward with velocity $u$ then

$h = u t_{2} + \frac{1}{2} g t_{2}^{2}$ ............(iii)

From (i)(ii) and (iii) we get

$- u t_{1} + \frac{1}{2} g t_{1}^{2} = \frac{1}{2} g t^{2}$ .................(iv)

$u t_{2} + \frac{1}{2} g t_{2}^{2} = \frac{1}{2} g t^{2}$ ..................(v)

Dividing (iv) and (v) we get

$∴ \frac{- u t_{1}}{u t_{2}} = \frac{\frac{1}{2} g (t^{2} - t_{1}^{2})}{\frac{1}{2} g (t^{2} - t_{2}^{2})}$ or $- \frac{t_{1}}{t_{2}} = \frac{t^{2} - t_{1}^{2}}{t^{2} - t_{2}^{2}}$

By solving $t = \sqrt{t_{1} t_{2}}$

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A stone dropped from the top of a building of height $h$ . It reaches the ground after $t s$ . From the same building if two stones are thrown (one upwards and other downwards) with the same velocity $u$ , and they reach the earth surface after $t_{1}$ and $t_{2}$ seconds respectively, then