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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:
`52m`
At position `A` balloon drops first particle So,
`u_(A)=0,a_(A)=-g,t=3.5sec.`
`S_(A)=((1)/(2) g t^(2))` ......`(i)`
Balloon is going upward from `A` to `B` in `2 sec`. So distance travelled by balloon in `2` second.
`(S_(B)=(1)/(2)a_(B)t^(2))` .......`(ii)`
`a_(B)=0.4m//s^(2), t=2 sec.`
`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=0.4xx2=0.8m//s`
`t=1.5sec.`
`(S_(2)=u_(2)t-(1)/(2)g t^(2))` ....`(iv)`

Distance between two stone
`DeltaS=S_(1)-S_(2)`.
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