A balloon is rising upwards from rest with acceleration `(g)/(8)`. A stone is dropped from the balloon when it is at height H. Show that the time by which the stone will touch the ground is 2 `sqrt((H)/(g))`.

Explanation

If the velocity of the balloon is v at a height H.
`v^(2) = 2(g)/(8) H" ""or",v = (1)/(2) sqrt(gH)`
The stone is dropped when the balloon is at height H. If we consider the downward direction to be positive, initial velocity of the stone,
So, if the stone touches the ground by time t,
`H = -(1)/(2)sqrt(gH).t+(1)/(2)"gt"^(2)" ""or","gt"^(2)-sqrt(gH).t-2H=0`
`:." " t = (sqrt(gH)pmsqrt(gH+8gH))/(2g)=(sqrt(gH)pm3sqrt(gH))/(2g)`
Neglecting the negative value of t
`t= (sqrt(gH)+3sqrt(gH))/(2g)=(4sqrt(gH))/(2g)=2sqrt((H)/(g))`