A stone is let to fall from a balloon ascending with acceleration `4(m)/(s^(2))`. After time `t=2s`, a second stone is dropped. If the distance between the stones after time `(t^(')=3s)` since the second stone is dropped is d, find `(d)/(28)`

Answer

4

Explanation

`s=(1)/(2)(g+a)(t+t^('))^(2)`
`s_(2)=(1)/(2)(g+a)(t^('))^(2)`
`s_(1)-s_(2)=d=(1)/(2)(g+a)(t)(t+2t^('))=(1)/(2)(14)(2)(2+6)`
`d=(12xx14)/(28)`
`(d)/(28)implies6`