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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
$4\sqrt {\frac{h}{g}} $
$2\sqrt {\frac{h}{g}} $
$\sqrt {\frac{{2h}}{g}} $
$\sqrt {\frac{g}{h}} $
Solution
(b) The velocity of balloon at height $h$, $v = \sqrt {2\left( {\frac{g}{8}} \right)h} $
When the stone released from this balloon, it will go upward with velocity $v =\frac{{\sqrt {gh} }}{2}$ (Same as that of balloon). In this condition time taken by stone to reach the ground
$t = \frac{v}{g}\left[ {1 + \sqrt {1 + \frac{{2g\,h}}{{{v^2}}}} } \right]$ $ = \frac{{\sqrt {g\,h} /2}}{g}\left[ {1 + \sqrt {1 + \frac{{2g\,h}}{{gh/4}}} } \right]$
$ = \frac{{2\sqrt {g\,h} }}{g} = 2\sqrt {\frac{h}{g}} $
