Question

A body released from a height falls freely towards earth. Another body is released one sec later. The separation between the bodies two sec after the released of the second body is (ans=24.5 m).

Answer 1

(i) Displacement of the first body after 3 seconds (because when 2nd body completes 2 seconds till then 1st would have completed 3 seconds)

Initial velocity for both balls will be zero since both are free dropped.

$s_1=ut+\frac{1}{2}g{t}^2$

$s_1=\left(0\right)+\frac{1}{2}\times g\times 3^2⟹s_1=4.5g$

(ii) Displacement of the second body after 2 seconds;

$s_2=ut+\frac{1}{2}g{t}^2$

$s_2=\left(0\right)+\frac{1}{2}\times g\times 2^2⟹s=2g$

So, separation between both the bodies,

$s_1-s_2=4.5g-2g=2.5g$

$s_1-s_2=2.5\times 9.8=24.5m$