A balloon rises from rest with a constant acceleration $g / 8$ . A stone is released from it when it has risen to height $h$ . The time taken by the stone to reach the ground is

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Solution

Upward velocity of stone at height $H$

$V^{2} = μ^{2} + 2 a s \Rightarrow v^{2} = 0 + 2 \frac{g}{8} \times H = \frac{H}{4} v = \frac{\sqrt{H g}}{2}$

Also,

$\to S = \to μ t + \frac{1}{2} a t^{2} H = \frac{\sqrt{H g}}{2} t + \frac{1}{2} g t^{2} g t^{2} - \sqrt{H g} - 2 H = 0 t^{2} - \sqrt{\frac{H}{g}} - \frac{2 H}{g} = 0 t = \frac{\sqrt{\frac{H}{g}} \pm \sqrt{\frac{H}{g} + \frac{8 H}{g}}}{2} = \frac{\sqrt{\frac{H}{g}} \pm 3 \sqrt{\frac{H}{g}}}{2} t = 2 \sqrt{\frac{H}{g}}$

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