Airborne

If the ball is released with initial speed $V$ at an angle of $\theta$ to the horizontal, the trajectory of the ball is $y \; = \; x\tan\theta - \tfrac{g}{2V^2}x^2\sec^2\theta \; = \; x\tan\theta - \tfrac{1}{2a}x^2\sec^2\theta$ where $a = \tfrac{V^2}{g}$ . Then the ball hits the ground again when $y=0$ , which happens when $x = 2a\sin\theta\cos\theta$ . Thus the length of the flightpath of the ball is $\begin{aligned} D(\theta) & = \; \int_0^{2a\sin\theta\cos\theta} \sqrt{1 + (y')^2}\,dx \; = \; \int_0^{2a\sin\theta\cos\theta}\sqrt{1 + \big(\tan\theta - \tfrac{1}{a}x\sec^2\theta\big)^2}\,dx \\ & = \; \frac{1}{a\cos^2\theta} \int_0^{2a\sin\theta\cos\theta}\sqrt{a^2\cos^4\theta + (x - a\sin\theta\cos\theta)^2}\,dx \; = \; \frac{1}{a\cos^2\theta}\int_{-a\sin\theta\cos\theta}^{a\sin\theta\cos\theta}\sqrt{a^2 \cos^4\theta + x^2}\,dx \\ & = \; a\cos^2\theta \int_{-\tan\theta}^{\tan\theta} \sqrt{1+y^2}\,dy \; = \; 2a\cos^2\theta\int_0^\theta \sec^3u\,du \\ & = \; 2a\cos^2\theta \times \tfrac12\left[\tan\theta\sec\theta + \ln(\sec\theta + \tan\theta)\right] \; = \; a\left[\sin\theta + \cos^2\theta\ln(\sec\theta + \tan\theta)\right] \end{aligned}$ and $D'(\theta) \; = \; a\left[\cos\theta + \cos\theta - 2\sin\theta\cos\theta\ln(\sec\theta+\tan\theta)\right] \; = \; 2a\cos\theta\big[1 - \sin\theta\ln(\sec\theta + \tan\theta)\big]$ Thus $D(\theta)$ will be maximized when $\sin\theta\ln(\sec\theta+\tan\theta) = 1$ (there is also a local minimum at $\theta = \tfrac12\pi$ , and a global minimum at $\theta = 0$ ). Solving this last equation numerically, we obtain $\theta = 0.98551473786$ , or $\boxed{56.4658^\circ}$ .

A projectile fired from level ground at an an angle $\theta$ with initial velocity $v$ and gravity $g$ follows the path of a parabola that reaches a maximum height of $\frac{v^2}{2g}\sin^2 \theta$ and a range of $\frac{2v^2}{g}\sin \theta \cos \theta$ (see here ). If we let the vertex of the parabolic path be the origin, then the coordinates of where the projectile lands would be $(\frac{v^2}{g}\sin \theta \cos \theta, -\frac{v^2}{2g}\sin^2 \theta)$ , which would make the equation of the parabolic path $y = -\frac{g}{2v^2}\sec^2 \theta x^2$ .

The length of the path can be expressed as $L = 2 \int_0^{\frac{v^2}{g}\sin \theta \cos \theta} \sqrt{1 + (\frac{dy}{dx})^2} \cdot dx$ $=$ $2 \int_0^{\frac{v^2}{g}\sin \theta \cos \theta} \sqrt{1 + \frac{g^2}{v^4}\sec^2 \theta x^2} \cdot dx$ . Using trig substitution with $\sec \alpha = \sqrt{1 + \frac{g^2}{v^4}\sec^2 \theta x^2}$ gives $L = \frac{2v^2}{g}\cos^2\theta \int_0^{\theta} \sec^3 \alpha \cdot d\alpha$ , and since $\int_0^{\theta} \sec^3 \alpha \cdot d\alpha = \frac{1}{2}(\tan \theta \sec \theta + \ln(\sec \theta + \tan \theta))$ this simplifies to $L = \frac{v^2}{g}\sin \theta + \frac{v^2}{g}\cos^2\theta \ln(\sec \theta + \tan \theta)$ .

The maximum length can occur when $\frac{dL}{d\theta} = \frac{2v^2}{g}\cos \theta(1 - \sin \theta \ln(\sec \theta + \tan \theta)) = 0$ , and in this case the maximum is length is when $1 = \sin \theta \ln(\sec \theta + \tan \theta)$ , which solves numerically to $\boxed{56.4658°}$ .