A projectile is fired at an angle of `45^(@)` with the horizontal. Elevation angle of the projection at its highest point as seen from the point of projection is

Options

A. `tan^(-1)((sqrt(3))/(2))`
B. `45^(@)`
C. `60^(@)`
D. `tan^(-1)((1)/(2))`

Answer

D

Explanation

`tan alpha = (H_(max))/(R//2) = (u^(2) sin^(2) theta//2 g)/(u^(2). 2sin theta cos theta//2g)`
`= (tan theta)/(2) =(tan 45^(@))/(2) = (1)/(2)`
`alpha = tan^(-1) (1//2)`