The maximum height reached by projectile is `4 m`. The horizontal range is `12 m`. The velocity of projection in `m s^-1` is (g is acceleration due to gravity)

Options

A. `5 sqrt(g//2)`
B. `3 sqrt(g//2)`
C. `(1)/(3) sqrt(g//2)`
D. `(1)/(5) sqrt(g//2)`

Answer

A

Explanation

(a) `tan theta = (4 H)/(R) =(4 xx 4)/(12) = (4)/(3) rArr sin theta = (4)/(5)`
`H = (u^2 sin^2 theta)/(2 g) rArr u = (sqrt(2 g H))/(sin theta) = (sqrt(2 xx g xx 4))/(4//5) = 5 sqrt((g)/(2))`.