Answer to Question #3627 in Mechanics | Relativity for biswajeet apata

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 isS1 = ut + gt2/2Here u = 0

g = acceleration due to gravity = 9.8 m/s2

So S1 = 0.5 gt2Distance of stone from ground is = h - S1 = h - 0.5 gt2 (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 S2 = ut - 0.5 g t2or 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

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Assignment Expert
31.10.17, 22:52


31.10.17, 20:10

( 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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