The velocity at the maximum height of a projectile is half of its velocity of projection `u`. Its range on the horizontal plane is

Options

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

Answer

B

Explanation

At maximum height `v=u cos theta`
`v = u cos theta`
`(u)/(2) =v rArr (u)/(2) = u cos theta`
`rArr cos theta = (1)/(2) rArr theta = 60^(@)`
`R=(u^(2) sin 2 theta)/(g)=(u^(2)sin (120^(@)))/(g)=(sqrt(3)u^(2))/(2g)`.