Question #3627

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?

Expert's answer

For upper stone which is dropped downwards the distance traveled isS_{1} = ut + gt^{2}/2Here u = 0

g = acceleration due to gravity = 9.8 m/s2

So S_{1} = 0.5 gt^{2}Distance of stone from ground is = h - S1 = h - 0.5 gt^{2} (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 S_{2} = ut - 0.5 g t^{2}or S_{2} = (8hg)1/2 t - 0.5 g t^{2} (2)Now for the two stones to meet

h - S1 = S2

or h - 0.5 g t^{2} = (8hg)1/2 t - 0.5 g t^{2} (from 1 and 2)or h = (8hg)1/2 t

or t = ( h/8g) 1/2

g = acceleration due to gravity = 9.8 m/s2

So S

Potential energy at top = Kinetic energy from the point of projection

or mg(4h) = 0.5 m u2

or u = 8hg/2

Thus S

h - S1 = S2

or h - 0.5 g t

or t = ( h/8g) 1/2

## Comments

Assignment Expert31.10.17, 23:52Yes

Max31.10.17, 21:10( h/8g)1/2 means - 'square root' of h/8g? or what? it make me confused. Please, make it clear it for me.

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