A projectile has horizontal range R and area A under its trajectory. Its maximum height H and launch angle θ satisfy:
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Question

A projectile has horizontal range R and area A under its trajectory. Its maximum height H and launch angle θ satisfy:

G1 Physics · Accepted Answer

Area under a projectile trajectory is A = 2RH/3. Hence H = 3A/(2R). Also tanθ = 4H/R. Therefore tanθ = 4/R × 3A/(2R) = 6A/R².