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Question

A particle is projected with a velocity v so that its range on a horizontal plane is twice the greatest height attained. Consider g=10 m/s2 as acceleration due to gravity. Then, its range will be:

A
4v25g
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B
4v25g2
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C
4v35g2
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D
4v5g2
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Solution

The correct option is A 4v25g
Range is given by
R=u2sin2θg
and maximum height is given by
H=u2sin2θ2g
and R=2H [Given in question]
u2sin2θg=2×u2sin2θ2g
2sinθcosθ=sin2θ
tanθ=2


From triangle ABC,
sinθ=25,cosθ=15

R=2u2sinθcosθg=2v2×25×15g
R=4v25g

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