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. `45^(@)`
B. `60^(@)`
C. `tan^(-1)(1)/(2)`
D. `tan^(-1)((sqrt3)/(2))`

Answer

C

Explanation

`tanalpha=h_(max)/(R//2)=((u^2sin^(2)45^(@))/(2g))/((u^2sin90^(@))/(2g))`
`tanalpha=1/4`
`alpha=tan^-1(1//4)`