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Asked by Prashant DIGHE | 29 Jan, 2020, 10:20: PM
Let t seconds be the time of collision after the particles started moving from top and bottom.

Vertical distance h1 travelled by particle which is dropped from the height h is given by,

h1 = (1/2)gt2 .....................(1)

Vertical distance h2 travelled by the partcle which is projected from ground with speed is given by,
h2 = () t - [ (1/2) g t2 ]  ..................(2)
from eqn.(1) and (2),  h1 + h2 = h = () t  .....................(3)
hence time of collision from starting time  is given by,    ...................(4)
hence from eqn.(1) and eqn.(4), we get  h1 = (1/2)g (h/2g) = (1/4)h

Hence collision has taken place at a height (3/4)h above ground level

Since collision is completely inelastic, all the kinetic energy is lost and the particles stick together.

Hence after collision, the combined partcles start from rest to reach ground .

If time taken is τ for the combined particle to reach ground , then we have , (3/4) h = (1/2) g τ2

hence time    ;  hence time τ in units of is
Answered by Thiyagarajan K | 30 Jan, 2020, 08:56: PM

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