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Question

The velocity at the maximum height of a projectile is 32 times its initial velocity of projection (u). Its range on the horizontal plane is

A
3u22g
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B
3u22g
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C
3u2g
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D
u22g
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Solution

The correct option is A 3u22g
As the vertical component of velocity becomes 0 at maximum height (H). Therefore, velocity at maximum height comprises only of horizontal component of velocity.
vH=ucosθ
According to question,
ucosθ=32u
θ=30
R=u2sin2θg=u2sin60g
=3u22g
Hence, the correct answer is option (a)

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