Q. 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
A) tan⁻¹(√3/2)
B) 45°
C) 60°
D) tan⁻¹(1/2)
Explanation
[Solution from source] At the top the projectile is at height H and horizontal distance R/2. tan α = H/(R/2) = (u² sin²θ/2g)/(u² sinθ cosθ/g) = tanθ/2 = 1/2 for θ = 45°. So α = tan⁻¹(1/2); 45° would be the launch angle, not the elevation of the highest point. Answer (4).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.