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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:

A
2 gh
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B
2gh
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C
gh
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D
2gh
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Solution

## The correct option is C √ghA 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 droppedso initial velocity u = 0Let's say t time they meetS = ut + 12 at²a = g ( acceleration due to gravity)Distance covered by stone A = 0 + 12 gt²Distance covered = h2 ( mid point)=> h2 = 12gt²=> h = gt²=> t² = hg=> t = √hgStone B initial velocity = va = -g ( as it is going against gravity)h2 = vt - 12gt²putting t = √hg=> h2 = v√hg - 12×ghg=> h = v√hg=> v = h√g√h=> v = √gh

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