Question

There are two identical particles A and B.one is projected vertically upward with speed root 2ghfrom ground and other is dropped from height h along the same verticle line.collision between them is pefectly inelastic .

Answer 1

The time taken by them to reach the earth is t = √3h / 2g

Explanation:

Time of collision "t" = h / √2gh

According to the first equation of motion:

v = u + at

V(A) = 0 - gt1 = - gh / √2gh

V(A) = - √gh / 2

V(B) = √2gh - gt1

V(B) = √2gh - gh / √2gh = √gh

Linear momentum conserved:

Pi = Pf

=> mVA +mvB = 2mVf

- √gh / 2 + √gh [ √2 - 1/√2] = 2Vf

Vf = 0

S = ut + 1/2 at^2

h - 1/2 gt1^2 = h - 1/2 g. h^2 / 2gh = 3h / 4

Time = 2s / g = 2 x 3h/4 / g

t = √3h / 2g

Thus the time taken by them to reach the earth is t = √3h / 2g

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