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

Explanation

Refer to Fig. 2 (d0. 45, let (u) the initial velocity or projectile at (o). If then follws a path (OAB), where ` AD = H = Maximum height.
Now, Max. height , ` H= (u^2 sin^2 45^@)/g = u^2/(4 g)`
Horzontal range,` R= (u^2 sin 2 xx 45^@)/g = u^2/g` ltBr. If ` alpha` is the angle of elevation of the projectile at the highest point from the point of projection, then ` /_AOD= alpha`.
.
From ` Delet AOD`, ltbRgt ` tab alpha = H/(R//2) = (2 H)/ R = ( 2( u^@ //4 g)/( u^2//g) = 1/2`
,. alpha = tan^(-1) (1/2)`.