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 ?

Options

A. 2.45 m
B. 24.5 m
C. 4.9 m
D. 9.8 m

Explanation

According to given problem 2nd body falls for 2 s so
that ` h_(2) = (1//2) g (B)^(2) ` …(A)
While `I^(st)` has fallen for 2 + 1 = 3 s,
so `h_(1) = (1/2) (C)^(2)` g … (B)
` therefore ` Separation between two bodies 2 s after the release of `II^(nd)` body
` d = h_(1) - h_(2) = (1//2) g (3^(2) - 2^(2)) = 4.9 xx 5 = 24.5 m `
Hence correct answer is (B)