A bolloon starts from rest moves vertically upwards with an acceleration `g//8 ms^(-2)` a stone falls from the balloon after 8 s from the start. Further time taken by the stone to reachg the ground `(g=9.8ms^(-2))` is

Explanation

The distance of the stone above the ground about which it begins to fall from the balloon is
`h=(1)/(2)((g)/(8))8^(2)=4g`
The velocity of the balloon at this height can be
obtained from `v=u+at,v=0+((g)/(8))=g`
This becomes the initial velocity (u) of the stone as the stone falls from the balloon at the height h
`:. u'=g`
For the total motion of the stone `h=-u' t+(1)/(2)g t^(2)`
`:. -4g=g t-(1)/(2)g t^(2)` and `t^(2)-2t-8=0`
Solving for `t` we get `t=4` and `-2s` ignoring negative value of time, `t=4s`