A stone 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 other downwards) with the same velocity $u$ and they reach the ground 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 B $t = \sqrt{t_{1} t_{2}}$

Step 1, Given data
Height of the building = h
Time taken to reach the ground = t
Two other balls are thrown in upward and downward directions simultaneously.
Step 2, Finding the relation
Taking downward as positive
For first case of dropping the stone, $h = \frac{1}{2} g t^{2}$
For second case of downward throwing $h = u t_{1} + \frac{1}{2} g t_{1}^{2} = \frac{1}{2} g t^{2}$
Or we can write
$⟹ u t_{1} = \frac{1}{2} g (t^{2} - t_{1}^{2})$ ......(i)
For third case of throwing upwards $h = - u t_{2} + \frac{1}{2} g t_{2}^{2} = \frac{1}{2} g t^{2}$
Or we can write
$⟹ u t_{2} = \frac{1}{2} g (t_{2}^{2} - t^{2})$ ....(ii)
On solving (i) and (ii) we get,
$\frac{t_{1}}{t_{2}} = \frac{t^{2} - t_{1}^{2}}{t_{2}^{2} - t^{2}}$
Or,
$⟹ t = \sqrt{t_{1} t_{2}}$
Hence the relation is as given above.

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