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# A stone is dropped from a height h, simultaneously another stone is thrown up from the ground which attains a maximum height 4h, the two stones cross each other at time ?

3 years ago

Arun
25768 Points

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 S=  (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

3 years ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions