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Question

# A stone A is dropped from rest from height h above ground. A second stone B is simultaneously thrown vertically up with velocity v. Find v so that B meets midway between their initial positions.

A
gh
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B
2gh
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C
3gh
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D
4gh
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Solution

## The correct option is A √ghLet time of travel of each stone =t Distance travelled by each stone =h2 Taking upward direction as positive Using second equation of motion, For stone A, −h2=−12gt2 ⇒t=√hg For stone B, h2=vt−12gt2 Putting t=√hg, we get v=√gh Alternate sol: For stone B w.r.t A, Srel=h, urel=v−0=v m/s, arel=0 Using second equation of motion, Srel=urelt+12at2 ⇒h=vt ⇒t=hv……(1) For the body dropped from height h, h2=12gt2⇒t=√hg……(2) From (1) and (2) √hg=hv⇒v=√hg

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