Introduction to Classical Mechanics · Problem 3.19

Using F = ma — Morin

Physics — Mechanics · projectile_trajectory

Problem

Maximum trajectory length *** A ball is thrown at speed vv from zero height on level ground. Let θ0\theta_0 be the angle at which the ball should be thrown so that the length of the trajectory is maximum. Show that θ0\theta_0 satisfies sin⁡θ0ln⁡(1+sin⁡θ0cos⁡θ0)=1.\sin \theta_0 \ln \left( \frac{1 + \sin \theta_0}{\cos \theta_0} \right) = 1. You can show numerically that θ0≈56.5∘\theta_0 \approx 56.5^\circ.
Intuitive explanation

The key insight is that trajectory length depends on both how high and how far the ball travels—and these compete as you change the launch angle. Unlike range (which peaks at 45°) or maximum height, arc length requires integrating ds=dx2+dy2ds = \sqrt{dx^2 + dy^2} along the parabolic path. Setting up this integral and optimizing it by requiring d(arc length)dθ=0\frac{d(\text{arc length})}{d\theta} = 0 yields a transcendental equation that can't be solved algebraically—but you can verify it numerically.

Intuitive explanation

The key insight is that trajectory length depends on both how high and how far the ball travels—and these compete as you change the launch angle. Unlike range (which peaks at 45°) or maximum height, arc length requires integrating ds=dx2+dy2ds = \sqrt{dx^2 + dy^2} along the parabolic path. Setting up this integral and optimizing it by requiring d(arc length)dθ=0\frac{d(\text{arc length})}{d\theta} = 0 yields a transcendental equation that can't be solved algebraically—but you can verify it numerically.

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