1
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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 √3u22gAs 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=u2sin60∘g =√3u22g Hence, the correct answer is option (a)

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