JEE MainPhysicsToughMotion in 1D

A balloon rises from rest on the ground with constant
acceleration $\left( \frac{g}{8} \right)$ . A stone is dropped when the
balloon has risen to a height H metre. Then, the time taken by the stone
to reach the ground is: (t=0 is the moment when stone is dropped)

Show answer and solution

v^{2} = 0 + 2 \times \frac{g}{8} \times H \Rightarrow v = \frac{\sqrt{gH}}{2}

- H = \frac{\sqrt{gH}}{2}t - \frac{1}{2}{gt}^{2} \Rightarrow t = \frac{\sqrt{gH} \pm 3\sqrt{gH}}{2g}

\Rightarrow t = 2\sqrt{\frac{H}{g}}

(taking positive value)

Source: Grass-G / P-05 Physics paper 10 October.docx

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