During a projectile motion if the maximum height equal the horizontal range, then the angle of projection with the horizontal is :

Options

A. `32^(@)`
B. `48^(@)`
C. `76^(@)`
D. `84^(@)`

Answer

C

Explanation

(c): `H=R` ,
`(u^(2) sin ^(2)theta)/(2g)=(u^(2) sin 2theta)/(g)`
`sin^(2)theta=2(2sin theta.cos theta)`
`sin theta=4cos theta`
`tan theta=4`
`theta=75.96`
`theta=76^(@)`