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

Answer

C

Explanation

Let u be the initial speed and `theta` is the angle of the projection.
Speed at the maximum height, `v_(H)=ucostheta=(sqrt(3))/(2)u`
`:.costheta=(sqrt(3))/(2)ortheta=cos^(-1)((sqrt(3))/(2))=30^(@)`
Range, `R=(u^(2)sin2theta)/(g)` and maximum height, `H=(u^(2)sin^(2)theta)/(2g)`
As R=PH(Given) `:.(u^(2)sin2theta)/(g)=P(u^(2)sin^(2)theta)/(2g)`
`2sinthetacostheta=(P)/(2)sin^(2)thetaortantheta=(4)/(P):.P=(4)/(tan30^(@))=4sqrt3`