The horizontal range and the maximum height of a projectile are equal. The angle of projection of the projectile is

Options

A. `theta = tan^(-1)(1/4)`
B. `theta = tan^(-1)(4)`
C. `theta = tan^(-1)(2)`
D. `theta = 45^@`

Answer

B

Explanation

Horizontal range , `R = (u^(2) sin 2 theta)/(g)`
where u is the velocity of projection and `theta` is the angle of projection
Maximum height , `H = (u^(2) sin^(2) theta)/(2g)`
According to question R = H
`:. (u^2sin 2theta)/g = (u^2sin^2 theta)/(2g),(2u^2sin thetacos theta)/g =(u^2sin^(2)theta)/(2g)`
`tan theta = 4 " or " theta = tan^(-1) (4)`