Two projectiles are fired with the same speed at complementary angles θ and 90°-θ. Their common range is R. The area enclosed between their two trajectories is:
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Two projectiles are fired with the same speed at complementary angles θ and 90°−θ. Their common range is R. The area enclosed between their two trajectories is:

G1 Physics · Accepted Answer

The two trajectories are y₁=x tanθ(1−x/R), y₂=x cotθ(1−x/R). Their vertical separation is |y₁−y₂| = x(1−x/R)|tanθ−cotθ|. Therefore the enclosed area is A = |tanθ−cotθ| ∫₀ᴿ x(1−x/R)dx. Now ∫₀ᴿ [x−x²/R]dx = R²/2−R²/3 = R²/6. Hence A = R²|tanθ−cotθ|/6.