For a fixed projection speed u, the area under the complete ground-to-ground projectile trajectory is maximum when the projection angle is:
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For a fixed projection speed u, the area under the complete ground-to-ground projectile trajectory is maximum when the projection angle is:

G1 Physics · Accepted Answer

Area under the trajectory is A = 2RH/3. Using R = u²sin2θ/g, H = u²sin²θ/(2g), we get A = [2u⁴/(3g²)] sin³θ cosθ. For maximum A, d/dθ [ln(sin³θ cosθ)] = 0. Thus 3cotθ − tanθ = 0 ⇒ tan²θ = 3. Hence θ=60°.