Q. 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
A) $4\sqrt{\frac{h}{g}}$
B) $2\sqrt{\frac{h}{g}}$
C) $\sqrt{\frac{2h}{g}}$
D) $\sqrt{\frac{g}{h}}$
Explanation
(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, $= \frac{\sqrt{gh}}{2}$ (Same as that of balloon). In this condition time taken by stone to reach the ground
$t = \frac{v}{g}\left\lbrack 1 + \sqrt{1 + \frac{2gh}{v^{2}}} \right\rbrack = \frac{\sqrt{gh}/2}{g}\left\lbrack 1 + \frac{2gh}{gh/4} \right\rbrack$
$= \frac{2\sqrt{gh}}{g} = 2\sqrt{\frac{h}{g}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.