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.

Explanation

`A` to `O` (motion under constant acceleration)
`v_(0)^(2)=0+2a_(1)h=2xxg/8h=(gh)/4`
`v_(0)=sqrt((gh)/4)=1/2 sqrt(gh)`
`O` to `A` via B (motion under gravity)
O is the origin and the upward direction is positive.
O to A : Displacement `=-h`
`s=v_(0)t-1/2g t_(2)`
`-h=v_(0)t-1/2 g t^(2)`
`g t^(2)=2v_(0)t-2h=0`
`t=(-(-2v_(0))+-sqrt((-2v_(0))^(2)-4(g)(-2)))/(2g)`
`=(2v_(0)+-sqrt(4v_(0)^(2)+8gh))/(2g)`
`=(2v_(0)+-sqrt(9gh))/(2g)=(sqrt(gh)+-sqrt(9gh))/(2g)`
`t=(sqrt(gh)+3sqrt(gh))/(2g)=4/(2g)sqrt(gh)=2sqrt(h/g)`
and t=-ve, not possible.