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

Given : Angle of projection θ=45°
Maximum height reached by the projectile H=u²sin²θ/2g=u²sin²45/2g=u²/4g (1)
Horizontal range of the particle R=u²sin2θ/g=u²sin90/g=u²/g
From figure, we get that at half of the horizontal range on the x-axis, the projectile reaches the maximum height on y axis.
∴We get R/2=u²/2g (2)
∴From figure
tanα=H/(R/2) = (u²/4g)/(u²/2g) = 1/2
∴α=tan⁻¹(1/2)