The centroid of the plane area bounded by a ground-to-ground projectile trajectory and the horizontal range is located at:
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The centroid of the plane area bounded by a ground-to-ground projectile trajectory and the horizontal range is located at:

G1 Physics · Accepted Answer

The trajectory is symmetric about x=R/2, so x̄ = R/2. For y = 4Hx(R−x)/R², the area is A = 2RH/3. The moment of the area about the x-axis is Mₓ = ∫₀ᴿ (y/2)y dx = (1/2)∫₀ᴿ y² dx. Now ∫₀ᴿ y² dx = 16H²R ∫₀¹ ξ²(1−ξ)² dξ = 16H²R(1/30) = 8H²R/15. Hence Mₓ = 4H²R/15. Therefore ȳ = Mₓ/A = [4H²R/15]/[2RH/3] = 2H/5. Thus the centroid is (R/2, 2H/5).