The total length S of a ground-to-ground projectile trajectory is:

Question

G1 Physics · Accepted Answer

The speed components are vₓ=u cosθ, vᵧ=u sinθ−gt. Hence path length is S = ∫₀ᵀ √[vₓ²+vᵧ²] dt. Let w=u sinθ−gt. Then dt=−dw/g. During the flight, w changes from u sinθ to −u sinθ. Thus S = (1/g)∫₋ᵘˢⁱⁿθ→ᵘˢⁱⁿθ √[(u cosθ)²+w²] dw. Using symmetry and the standard integral, S = (u²/g) [sinθ + cos²θ ln((1+sinθ)/cosθ)]. Since (1+sinθ)/cosθ = secθ+tanθ, we obtain S=(u²/g)[sinθ+cos²θ ln(secθ+tanθ)].