The maximum height attained by a projectile when thrown at an angle `theta` with the horizontal is found to be half the horizontal range. Then `theta` is equal to

Options

A. `tan^(-1) (2)`
B. `pi/6`
C. `pi/4`
D. `tan^(-1)(1/2)`

Answer

A

Explanation

Minimum height,`H_0 = (u^2 sin theta)/(2g)`
Range, `R = (u^2 sin 2 theta)/(g)`
Given, `H_0 = R/2`
`therefore (u^2 sin^2 theta)/(2g) = (u^2 2 sin theta cos theta)/(2 g)`
`implies sin theta = 2 cos theta`
`implies tan theta = 2`
`therefore theta = tan^(-1) (2)`