A stone dropped from a stationary balloon which is at height $h$ reaches the ground after $t s$ From the same balloon if two stones are thrown, one vertically upwards and the other vertically downwards with same speed $u$ and they reach the ground after $t_{1} s$ and $t_{2} s$ respectively, then

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Solution

For first case of dropping from balloon, initial velocity of stone is zero.

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

If thrown upward with speed $u$ , then-

$- h = u t_{1} - \frac{1}{2} g t_{1}^{2}$ taking upward direction as positive --------(2)

And for throwing down-

$- h = - u t_{2} - \frac{1}{2} g t_{2}^{2}$ ---------------(3)

Now $(2) - (3)$ --

$0 = u (t_{1} + t_{2}) - \frac{g}{2} (t_{1}^{2} - t_{2}^{2})$

$⟹ u (t_{1} + t_{2}) = \frac{g}{2} (t_{1} + t_{2}) (t_{1} - t_{2})$

$⟹ u = \frac{g}{2} (t_{1} - t_{2})$

From $(3)$ -

$h = u t_{2} + \frac{1}{2} g t_{2}^{2} = \frac{1}{2} g t_{1} t_{2}$

$⟹ \frac{1}{2} g t^{2} = \frac{1}{2} g t_{1} t_{2}$

$⟹ t = \sqrt{t_{1} t_{2}}$

Hence, $h = \frac{1}{2} g t^{2} = \frac{1}{2} g t_{1} t_{2}$

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