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 =`

Options

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

Answer

A

Explanation

Maximum height `H_(0) = (u^(2)sin^(2)theta)/(2g)`
Range, `R = (u^(2)sin2theta)/(g)`
`therefore H_(0) = (R)/(2) or (u^(2)sin^(2)theta)/(2g) = (u^(2) xx 2sintheta cos theta)/(2 xx g)`
or `sin theta = 2cos theta` or `tan theta = 2`
or `theta = tan^(-1)2`