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Question

A body is projected with a velocity u so that the horizontal range is twice the maximum height. Then the maximum height is

A
u22g
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B
u2g
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C
5u24g
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D
2u25g
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Solution

The correct option is B 2u25g
According to question, horizontal range is twice the maximum height
R=2H
u2sin2θg=2u2sin2θ2g
2sinθ.cosθ=sinθ.sinθ
tanθ=2
Then, sinθ=25, cosθ=15
Maximum height in projectile motion,
H=u2sin2θ2g
H=u22g×(25)2
H=2u25g
The correct option is D.

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