If an object is thrown at an angle of 60° with horizontal, find the elevation angle of the object at its highest point as seen from the point of projection.

The maximum height upto which the object goes, H = u²y²/g = u²sin²θ/2g
The range of the motion, R = 2uxuy/g = 2u²sinθcosθ/g
Hence the angle of elevation = tan⁻¹(H/(R/2)) = tan⁻¹(tanθ/2) = tan⁻¹(tan60°/2) = tan⁻¹(√3/2)