Question

A body projected upward with a velocity 100m/s from ground.At a same time another body is released from height of 300m. When and where do they meet?

Answer 1

3 s after the bodies start moving, 45 m below the top of the tower.

Explanation:

The relative acceleration of the two bodies = 0 (both particles have equal acceleration)

The relative velocity of approach #= (300" m")/(100" ms"^-1) = 3 " s"#

In this time, the body dropped from the top would have fallen by `1/2 times 10" ms"^-2 times (3" s")^2 = 45" m"`