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A balloon a ascending vertically with an...

A balloon a ascending vertically with an acceleration of `0.4m//s^(2)`. Two stones are dropped from it at an interval of `2 sec.` Find the distance between them `1.5sec.` after the second stone is released. `(g=10m//sec^(2))`.

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The correct Answer is:
52 m

At position `A` ballon drops first particle
So, `u_(A) =0,a_(A) = -g, t = 3.5 sec`
`S_(A)=((1)/(2)g t^(2))`.....(i)
Balloon is going upward from `A` and `B` in `2 sec`. So distance travelled by balloon in `2` second.
`(S_(B)=(1)/(2)a_(B)t^(2))` .....(ii)
`a_(B)=0.4 m//s_(2),t=2sec`
`S_(1)=BC=(SB+SA)`....(iii)
Distance travell by second stone which is droped from balloon at `B`
`u_(2)=u_(B)=a_(B)t=0.4 xx2 =0.8 m//s`
`t=1.5 sec`.
`(S_(2)=u_(2)t-(1)/(2)g t^(2))` .....(iv)
Distance between two stone
`Delta S=S_(1)-S_(2)`.
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