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Question

From the elevated point P a stone is projected vertically upward. When it reaches a distance y below the point of projection its velocity is tripled the velocity when it was at a height y above P. The greatest height reached by it above P 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 5y3
Let the initial velocity = u
At point 'P'
AP=hV2pu2=2gh.......(i)
At point 'Q'
V2Qu2=2(g)(h)V2Qu2=2gh
As VQ=2VP
4V2Pu2=2gh........(ii)
4[u22gh]u2=2gh
4u28ghu2=2gh
3u2=10gh
u2=103gh.......(iii)
But 02u2=2gh........(iv)
2gH=103gh
2H=103h
H=53h
According to the question height is 'y' so
H=53y

1064879_1024489_ans_c4762d166777416eba3fa9559c57158e.png

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