Question

One body drop from top and another body thrown up simultenously

Answer 1

so what is your doubt ?what do u want 2 no?

Answer 2

 

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,



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.