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Question

A stone A is dropped from a height h above the ground. A second stone B is simultaneously thrown vertically up from a point on the ground with velocity v. The line of motion of both the stones is same. The value of v which would enable the stone B to meet the stone A midway ( at midpoint) between their initial position is:

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Solution

The correct option is **C** √gh

A stone A is dropped from rest from a height h above the ground. A second stone B is simultaneously thrown vertically up from a point on the ground with velocity v. The line of motion of both the stones is same.

Stone A is dropped

so initial velocity u = 0

Let's say t time they meet

S = ut + 12 at²

a = g ( acceleration due to gravity)

Distance covered by stone A = 0 + 12 gt²

Distance covered = h2 ( mid point)

=> h2 = 12gt²

=> h = gt²

=> t² = hg

=> t = √hg

Stone B initial velocity = v

a = -g ( as it is going against gravity)

h2 = vt - 12gt²

putting t = √hg

=> h2 = v√hg - 12×ghg

=> h = v√hg

=> v = h√g√h

=> v = √gh

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