The speed of a projectile at its maximum height is `sqrt3//2` times its initial speed. If the range of the projectile is n times the maximum height attained by it, n is equal to :

Options

A. `4/3`
B. `2sqrt3`
C. `4sqrt3`
D. `3/4`

Answer

B

Explanation

Given , `(sqrt(3u))/2 = u cos theta` = speed at maximum height
`:.cos theta = sqrt3/2 " or " theta = 30^(@)`
Given `PH_("max") = R " " :. " " P = R/(H_(max))`
But `H_(max) = (Rtan theta)/(4)`
`:. P = 4/(tan theta) = 4/(tan 30^(@)) = 4sqrt3`