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(B) The balloon starts from rest with acceleration $a = g/8$. At height $h$,its velocity $v$ is given by $v^2 = u^2 + 2ah = 0 + 2(g/8)h = gh/4$. Thus,

Vedclass · Accepted Answer

$2\sqrt{\frac{h}{g}}$

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(B) The balloon starts from rest with acceleration $a = g/8$. At height $h$,its velocity $v$ is given by $v^2 = u^2 + 2ah = 0 + 2(g/8)h = gh/4$. Thus,$v = \frac{1}{2}\sqrt{gh}$.When the stone is released,it possesses an initial upward velocity $v = \frac{1}{2}\sqrt{gh}$.Using the equation of motion for the stone: $s = ut + \frac{1}{2}at^2$,where $s = -h$ (downward displacement),$u = v = \frac{1}{2}\sqrt{gh}$,and $a = -g$.$-h = (\frac{1}{2}\sqrt{gh})t - \frac{1}{2}gt^2$.Rearranging gives $\frac{1}{2}gt^2 - (\frac{1}{2}\sqrt{gh})t - h = 0$.Multiplying by $2/g$: $t^2 - (\sqrt{h/g})t - 2h/g = 0$.Using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$t = \frac{\sqrt{h/g} \pm \sqrt{h/g + 8h/g}}{2} = \frac{\sqrt{h/g} \pm 3\sqrt{h/g}}{2}$.Taking the positive root,$t = \frac{4\sqrt{h/g}}{2} = 2\sqrt{\frac{h}{g}}$.

$A$ 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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