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. `2 sqrt(3)`
C. `4 sqrt(3)`
D. `3//4`

Answer

C

Explanation

( c) Given `(sqrt(3) u)/(2) = u cos theta` = speed at maximum height
or `cos theta = (sqrt(3))/(2)` or `theta = 30^@` …(i)
Given that `PH_(max) = R`…(ii)
We know `H_(max) = (R tan theta)/(4)`
`P = (R )/(H_(max)) = (4)/(tan theta) = (4)/(tan 30^@) = 4 (sqrt(3))`.