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

Options

A. `sqrt(3)(u^(2))/(2g)`
B. `(u^(2))/(2g)`
C. `(3u^(2))/(2g)`
D. `(3u^(2))/(g)`

Answer

A

Explanation

`u cos theta=sqrt(3)/(2)u rArr theta=30^(@)`
`rArr R=(u^(2) sin 2 xx30^(@))/(g)=(u^(2)sin 60^(@))/(g)=sqrt(3)(u^(2))/(2g)`