A body is released from a great height and falls freely towards the earth. Exactly one sec later another body is released. What is the distance between the two bodies 2 sec after the release of the second body ?

Explanation

According to given problem, second body fails for 2 s, so that,
`h_(2)=(1)/(2)g(2)^(2)`
while 1" has fallen for 2 + 1 = 3-sec:
`h_(1)=(1)/(2)g(3)^(2)`
Separation between two bodies 2 s after the release of 2 body `d=h_(1)-h_(2)-(1)/(2)-(1)/(2)g(3^(2)-2^(2))=4.9xx5=24.5m`