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?
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