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Question

# A stone A is dropped from rest from height h above the ground. A second stone B is simultaneously thrown vertically up from a point on the ground with velocity v. Find v so that B meets A 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 Using second equation of motion, Taking upward as positive for both stones, For stone A,−h2=−12gt2⇒t=√hg For stone B,h2=vt−12gt2 Putting t=√hg, we get h2=v√hg−12g(√hg)2 v=√gh Alternate Solution: For stone B w.r.t A, Srel=SB−SA=h2−(−h2)=h urel=v−0=v m/s and arel=0 Using second equation of motion, Srel=urelt+12at2 ⇒h=vt ⇒t=hv ... (1) For stone A: h2=12gt2⇒t=√hg ... (2) From (1) & (2), √hg=hv⇒v=√gh

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