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. (a) ` 4 sqrth/g`
B. (b) ` 2 sqrt h/g`
C. (c ) sqrt (2 h) /g`1
D. (d) ` sqrt g/h`

Answer

B

Explanation

Velocity acquired by ballon at heitht (h),
` v= sqrt( 2 ah) = = sqrt 92 xx g/8 h) = (sqrt gh)/2`
Taking downward motion of stone released from balloon at height (h) upto surface of ground, we have
` u= - (sqrt gt)/2 , a=g , S=h, t=t`
As ` S= ut^2`
:. ` h= - 9sqrt gh) /2 t + 1/2 gt^2 or ` t^2 - sqrt h/g t- (2 h) /g =0`.
On solving `, t=2 sqrt h//g ` or - sqrth/g`
As negative tiem is not accounted , hence
` t= 2 sqrt h//g`