A stone is let to fall from a balloon ascending with an acceleration `f`. Aftre time `t`. A second stonr is dropped. Prove that the distance between the stones after time `t'`
since the second stone is dropped, is `(1)/(2) (f+g)t(t+2t')`.

Answer

`(1)/(2) (g+f) (t) (t+2t')`

Explanation

`S_(1)=(1)/(2) (g+f) (t+t')^(2) , S^(2) =(1)/(2) (g=f) t'^(2)`
`S=S_(1) -S^(2) =(1)/(2) (g+f) (t) (t+2t')`
`S=S_(1)-S_(2) =(1)/(2) (g+f)(t) (t+2t')`.