Question

A stone is dropped from a height h . Simulataneously another stone is thrown up from the ground which reaches a height 4h. The teo stones will cross each other aftrr time

Answer 1

Answer:

Explanation:

For upper stone which is dropped downwards the distance traveled is

S1 = ut + 1/2 g t^2

Here u = 0

g = accln due to gravity = 9.8 m/s2

So S1 = 1/2g t2

Distance of stone from ground is = h - S1 = h - 0.5 g t2 ....(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) = 1/2 m u2

or u = (8hg)1/2

Thus S2 = ut - 0.5 g t2

or 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