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Question

A body is released from top of tower. At the same instant another body is projected vertically with velocity 'u' .
Find the time after which both bodies will meet ?

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Solution

Let both the particles meet after time, t .

Hence,

The distance covered by the particle coming down from the top of the tower, would be equal to:

s=12gt2......(1)

The distance covered by the particle going from ground would be given by:

hs=ut12gt2......(2)

From (1) and (2) the time taken by both to meet, would be given by:

t=hu

Hence, the time taken by both to meet is t=hu


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