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Question

A particle is projected with a velocity v such that its range on horizontal plane is twice the greatest height attained by it. The range of the projectile will be (g is the acceleration due to gravity).

A
4v25g
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B
4g5v2
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C
v2g
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D
4v25g
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Solution

The correct option is A 4v25g
From the given condition of the projectile we have
R=2H
Where, R is the range of the projectile and H is the height of the projectile.

We know that, R=4H cot θ cot θ=2H4H=12
Plotting the value of cot θ on triangle we get


sin θ=25,cos θ=15
Range of projectile is R=2v2 sin θ cos θg
=2v2g×25×15=4v25g.

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