The horizontal range of a projectile is times of its maximum height. Its angle of projection will be :-

Options

A. `45^(@)`
B. `60^(@)`
C. `90^(@)`
D. `30^(@)`

Answer

D

Explanation

`R=(u^(2) sin 2 theta ) /(g) and H=(u^(2) sin ^(2) theta )/( 2g) `
but ,` R= 4 sqrt(3 ) H `
`(u^(2) sin 2 theta ) /(g) = 4 sqrt(3) ((u^(2) sin ^(2) theta )/(2g))`
` sin 2 theta = 4 sqrt(3)(sin^(2)theta )/(2)`
`2 sin theta cos theta =(4sqrt(3) sin ^(2)theta )/(2)`
` (1)/(sqrt(3))-(sin theta )/(cos theta ) = tan theta `
or `theta = 30^(@)`