Question

A ballon is ascending vertically with an acceleration fo `0.2m//s^(2)`. Two stones are dropped from it at an interval of `2sec`. Find the distance between them `1.5sec` after the second stone is released. (use `g =9.8 m//s^(2))`

Answer 1

Correct Answer - `50m`
from the frame attached to ballon
`h_(1) = (1)/(2) (g+a) ((7)/(2))^(2)` & `h_(2) = (1)/(2) (g+a) ((3)/(2))^(2)`
separation between
`(h_(2) -h_(1)) = (1)/(2) xx 10 ((49)/(4) - (9)/(4)) = 50 m`