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

Let `phi` be elevation angle of the projectile at its highest point as seen from the point of projection O and `theta` be angle of projection with the horizontal.
From figure,
`tan phi = H/(R//2)" "....(i)`
In case of projectile motion
Maximum height , `H=(u^2 sin^(2) theta)/(2g)`
Horizontal range, `R = (u^2 sin 2 theta)/(g)`
Substituting these values of H and R in (i) , we get
`tan phi=(sin^2theta)/(sin2theta)=(sin^2theta)/(2sin thetacos theta)=1/2tan theta`
Here , `theta = 45^@`
`:. tan phi =1/2 tan 45^(@) =1/2 , phi = tan^(-1)(1/2)`