An aeroplane is rising vertically with acceleration `f`. Two stones are dropped from it at an interval of time `t`. The distance between them at time t' after the second stone is dropped will be:

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))

A balloon in ascending vertically with an acceleration of 1 ms^(-2) . Two stones are dropped from it at an interval of 2 s . Find the distance berween them 1.5 s after the second stone is released.

A balloon in ascending vertically with an acceleration of 1 ms^(-2) . Two stones are dropped from it at an interval of 2 s . Find the distance berween them 1.5 s after the second stone is released.

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)) .

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)) .

A balloon is ascending vertically with an acceelration of 0.2 ms^(-2) .Two stones are dropped from it at an interval fo 2 s , the distance between then when the second stone dropped is (tanke g=9.i8 ma^(-2).

A stone is let to fall from a balloon ascending with an acceleration f . After time t , a second stone 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') .

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') .