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

Answer

D

Explanation

Given, `u cos theta = (u)/(2) rArr theta = 60^(@)`
Now, `R = (u^(2) sin 2theta)/(g) = (u^(2)sin 120^(@))/(g) = (sqrt(3)u^(2))/(2g)`