Question

A stone is dropped from a height h. Simultaneously another stone is thrown up from the ground with such a velocity that it can reach a height 4h. Find the time when two stone cross each other?

Answer 1

For upper stone which is dropped downwards the distance traveled isS₁ = ut + gt²/2Here u = 0

g = acceleration due to gravity = 9.8 m/s2

So S₁ = 0.5 gt²Distance of stone from ground is = h - S1 = h - 0.5 gt² (1)Now for stone thrown upwards such that it goes to height 4h from ground
Potential energy at top = Kinetic energy from the point of projection

or mg(4h) = 0.5 m u2
or u = 8hg/2

Thus S₂ = ut - 0.5 g t²or S₂ = (8hg)1/2 t - 0.5 g t² (2)Now for the two stones to meet

h - S1 = S2
or h - 0.5 g t² = (8hg)1/2 t - 0.5 g t² (from 1 and 2)or h = (8hg)1/2 t
or t = ( h/8g) 1/2