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Question

# From the elevated point â€²Pâ€² a stone is projected vertically upward. When it reaches a distance â€²yâ€² below the point of projection its velocity is tripled the velocity when it was at a height â€²yâ€² above â€²Pâ€². The greatest height reached by it above â€²Pâ€² is:

A
5y4
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B
5y3
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C
y3
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D
2y
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Solution

## The correct option is B 5y3Let the initial velocity = uAt point 'P'AP=h⇒V2p−u2=−2gh.......(i)At point 'Q'V2Q−u2=2(−g)(−h)⇒V2Q−u2=2ghAs VQ=2VP4V2P−u2=2gh........(ii)4[u2−2gh]−u2=2gh4u2−8gh−u2=2gh3u2=10ghu2=103gh.......(iii)But 02−u2=−2gh........(iv)2gH=103gh2H=103hH=53hAccording to the question height is 'y' soH=53y

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