Question

A balloon rises from rest on the ground with a constant acceleration of g/g . A stone is dropped, from the balloon when it has risen to a height of H. The time taken by the stone to reach the ground is

(1) \(\sqrt{\frac{H}{g}}\) 

(2) \(2\sqrt{\frac{H}{g}}\)

(3) \(\sqrt{\frac{2H}{g}}\)

(4) \(\sqrt{\frac{H}{{2g}}}\)

Answer 1

 (2) \(2\sqrt{\frac{H}{g}}\)

Let v be the velocity of the balloon, when it has risen to the height H. Then

This v will be the initial velocity of the stone when dropped from the balloon. If t is the time taken by the stone to reach the ground, then