A stone is dropped from a building of height $h$ and it reaches after $t$ seconds on earth. 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 height $h$ , then $h = \frac{1}{2} g t^{2} . . . (i)$
If a stone is thrown upwards with velocity $u$ , then $h = - u t_{1} + \frac{1}{2} g t_{1}^{2} . . . (i i)$
If a stone is thrown downwards with velocity $u$ , then $h = u t_{2} + \frac{1}{2} g t_{2}^{2} . . . (i i i)$
From $(i)$ $(i i)$ and $(i i i)$ , we get
$- u t_{1} + \frac{1}{2} g t_{1}^{2} = \frac{1}{2} g t^{2} . . . (i v)$
$u t_{2} + \frac{1}{2} g t_{2}^{2} = \frac{1}{2} g t^{2} . . . (v)$
Dividing $(i v)$ 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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