A body is projected with velocity u at angle θ to horizontal. The horizontal range is maximum when θ is: MCQ with Answer and Explanation

Horizontal range R = u²sin(2θ)/g. sin(2θ) is maximum (value 1) when 2θ = 90° ⇒ θ = 45°. Thus maximum range at 45° projection angle, assuming same initial speed and level ground. Memory tip: 'Max range at 45°; max height at 90°'. This standard projectile motion result is frequently tested in competitive exams. Always verify assumptions: no air resistance, uniform gravity, launch and landing at same height. For different heights, optimal angle differs.