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))`

Options

A. `(1)/(2) t ((g + f)/(t'))`
B. `(1)/(2) t ((g + f)/(t'+ 2t'))`
C. `(1)/(2) t ((g + f)/(t'+ t'))`
D. `(1)/(2) t ((g t f)/(t'+ t'))`

Answer

B

Explanation

Let the velocity of aeroplane at the time when 1 st and 2 nd stone were dropped was `u` and `u'` then distance travelled will be:
`h_(1) = u (t + t') - (1)/(2) g (r + t')^(2)`
`h_(2) = u' t' - (1)/(2) g t'^(2)`
`Delta h = h_(1) - h_(2)`