The Orbits' Envelope of the Projectiles

The trajectory for a particle projected at an angle of $\theta$ to the horizontal is $y \;=\; x\tan\theta - \tfrac{g}{2v^2}x^2\sec^2\theta$ Let us write $d = \tfrac{v^2}{g}$ . Then the maximum height that can be reached by the particle (when $\theta = \tfrac12\pi$ ) is $\tfrac12d$ , and the maximum range of the particle (when $\theta = \tfrac14\pi,\tfrac34\pi$ ) is $d$ . If we consider the curve $C$ given by the equation $y \; = \; \tfrac{1}{2d}(d^2-x^2)$ then $\tfrac1{2d}(d^2 - x^2) - \big(x\tan\theta - \tfrac{1}{2d}x^2\sec^2\theta\big) \; = \; \tfrac12d - x\tan\theta + \tfrac{1}{2d}x^2\tan^2\theta \; = \; \tfrac{1}{2d}\big(x\tan\theta - d\big)^2$ Thus the curve $C$ always lies above every trajectory curve, and touches the $\theta$ trajectory at the point $x = d\cot\theta$ . In other words, $C$ is the envelope of the trajectories. The area between $C$ and the $x$ -axis is easy to calculate as $\tfrac23d^2$ , and hence $\alpha = \boxed{\tfrac23}$ .

The trajectory of a projectile with fixed initial speed but variable angle is given by

$y_{traj}=x\tan\theta - \frac{g}{2v^2\cos^2\theta}x^2$

The envelope of the family of trajectories are the set of all points reachable by all the trajectories, and must satisfy the condition that the envelope function $y(x)$ must be tangent to and contain the point (x,y) at all points that intersect with a trajectory.

This is satisfied when $F(x,y,\theta) = x\tan\theta - \frac{g}{2v^2\cos^2\theta}x^2-y = 0$ and $\frac{\partial}{\partial\theta}F(x,y,\theta) = x\sec^2\theta - \frac{g}{v^2\cos^2\theta}x^2\tan\theta= 0$ .

Combining the two gives the envelope function to eliminate terms with $\theta$ gives

$y=\frac{v^2}{2g}\left(1-\frac{g^2x^2}{v^4}\right)$ .

We recognize $h=\frac{v^2}{2g}$ and $R = \frac{v^2}{g}$ as maximum height and maximum range.

If we go to unitless so that $v = y/h$ and $u=x/R$ and agree to scale areas by a factor $hR=\frac{v^4}{2g^2}$ then the area required by the question is easily found by:

$Area = hR \int_{-1}^{+1} (1-u^2) du = hR(2-2/3)=4/3hR = \frac{2}{3}\frac{v^4}{g^2}$

Thus, the answer is $2/3$