A projectile is fired at an angle of `45^(@)` with the horizontal. Elevation angle of the projectile 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)(sqrt(3)/(2))`

Answer

C

Explanation

`tan theta=((H)/(R))/(2)=(u^(2)sin^(2)45^(@))/((2g)/((2u^(2)sin45^(@)cos 45^(@))/(2g)))=(tan45^(@))/(2)=(1)/(2)`
`rArr theta=tan^(-1)((1)/(2))`