A body released from a great height, falls freely towards the earth. Another body is released from the same height exactly one second later. Then the separation between two bodies, two seconds after the release fo second body is.

Explanation

According to given problem 2 nd body falls for 2 s so that
`h_(2) = (1)/(2) g (2)^(2)`.....(i)
while 1 st has fallen for `2 + 1 = 3 s`, so
`h_(1) = (1)/(2) g (3)^(2)` ....(ii)
`:.` Separation between two bodies 2 s after the release of 2nd body,
`d = h_(1) - h_(2)`
`= (1)/(2) g (3^(2) - 2^(2))`
`= 4.9 xx 5`
`=24.5 m`