Library · Short Excursions · Physics and computation

Maximising the Length of a Projectile Trajectory

At what angle should a projectile be launched, under uniform gravity and no air resistance, so that the arc length of its trajectory is maximised? A standard integral and one implicit equation give the answer.

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Under uniform gravity and no drag, the arc length of a unit-speed projectile launched at angle θ\theta is L(θ)=12g[2sin⁡θ+cos⁡2θlog⁡1+sin⁡θ1−sin⁡θ]L(\theta) = \tfrac{1}{2g}\left[2\sin\theta + \cos^2\theta \log\frac{1+\sin\theta}{1-\sin\theta}\right]. Differentiating and setting to zero gives 2csc⁡θ=log⁡1+sin⁡θ1−sin⁡θ2\csc\theta = \log\frac{1+\sin\theta}{1-\sin\theta}, numerically θ≈56.465∘\theta \approx 56.465^\circ. Full setup in the PDF.

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