A balloon rises from rest on the ground with constant acceleration `g`//`8.` A stone is dropped from the balloon when the balloon has risen to a height of (H). Find the time taken by the stone to reach the ground.

Options

A. `4sqrt(h//g)`
B. `2sqrt(h//g)`
C. `sqrt(2h//g)`
D. `sqrt(g//h)`

Answer

B

Explanation

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