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The velocity at the maximum height of a ...

The velocity at the maximum height of a projectile is `(sqrt(3))/(2)` times its initial velocity of projection (u). Its range on the horizontal plane is

A

`(sqrt(3)u^(2))/(2g)`

B

`(u^(2))/(2g)`

C

`(3u^(2))/(2g)`

D

`(3u^(2))/(g)`

Text Solution

Verified by Experts

The correct Answer is:
A
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