A stone is dropped from a height h.simultaneously another stone is thrown up from the ground which reaches the height 4h.the two stones cross each other after a time-------------

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

I thnk this could help u...

With Regards,

RONIT MOHANTY^®

initial vel of stone thrown up from ground v(say)

0²-v²=-2g(4h)

=>v=√(8gh)

since both are droped symultameously

hence relative accleration is 0

so vel. of thrown up will seem constant to vel to the dropped one

hence time=dist/speed=h/√(8gh)=√(h/8g) ans.