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`
`u/2 = vrArr cos theta =1/2 rArr theta =60^(@)`
`R=(u^(2)sin 2 theta)/g`
`=(u^(2) sin(120^(@)))/g`
`=(u^(2) cos 30^(@))/g=(sqrt(3) u^(2))/(2g)`