+30 A projectile is fired with an initial velocity of $v$ at an angle of $\theta$ from the ground. Then its trajectory can modeled by the parametric equations \begin{align*} x &= vt \cos \theta, \\ y &= vt \sin \theta - \frac{1}{2} gt^2, \end{align*}where $t$ denotes time and $g$ denotes acceleration due to gravity, forming a parabolic arch. Suppose $v$ is held constant, but $\theta$ is allowed to vary, over $0^\circ \le \theta \le 180^\circ.$ The highest point of each parabolic arch is plotted. (Several examples are shown below.) As $\theta$ varies, the highest points of the arches trace a closed curve. The area of this closed curve can be expressed in the form \[c \cdot \frac{v^4}{g^2}.\]Find $c.$
To find the highest point of the parabolic arch, we need to find the value of $t$ that maximizes $y.$ Since the coefficient of $t^2$ is negative, the highest point occurs when the quadratic term is zero, that is, when $t = \frac{v \sin \theta}{g}.$ Substituting this value of $t$ into the equation for $y,$ we get:
$y_{\text{max}} = \frac{v^2 \sin^2 \theta}{2g}$
This expression gives the maximum height of the projectile for a given angle $\theta.$ Note that the height only depends on the square of the initial velocity and the sine of the angle, not on the actual value of the velocity or the angle itself.
To trace out the curve of maximum heights as $\theta$ varies, we need to find the maximum height for each angle. We can express this maximum height as a function of $\theta$:
$f(\theta) = \frac{v^2 \sin^2 \theta}{2g}$
The curve of maximum heights is then given by the graph of $f(\theta)$ over the interval $0^\circ \leq \theta \leq 180^\circ.$ Note that $f(\theta)$ is a symmetric function with respect to $\theta = 90^\circ,$ since $\sin^2 \theta = \sin^2 (180^\circ - \theta).$ Therefore, we can restrict our attention to the interval $0^\circ \leq \theta \leq 90^\circ.$
To find the area enclosed by the curve of maximum heights, we need to integrate $f(\theta)$ over the interval $0^\circ \leq \theta \leq 90^\circ.$
$\begin{aligned} \text{Area} &= \int_{0^\circ}^{90^\circ} f(\theta) d\theta \ &= \int_{0^\circ}^{90^\circ} \frac{v^2 \sin^2 \theta}{2g} d\theta \ &= \frac{v^2}{2g} \int_{0^\circ}^{90^\circ} \sin^2 \theta d\theta \ &= \frac{v^2}{4g} \int_{0^\circ}^{180^\circ} \sin^2 \theta d\theta \ &= \frac{v^2}{4g} \int_{0^\circ}^{180^\circ} \frac{1 - \cos 2\theta}{2} d\theta \ &= \frac{v^2}{8g} \int_{0^\circ}^{360^\circ} (1 - \cos 2\theta) d\theta \ &= \frac{v^2}{8g} \left[\theta - \frac{1}{2} \sin 2\theta \right]_{0^\circ}^{360^\circ} \ &= \frac{v^2}{8g} \cdot 360^\circ \ &= \frac{45}{\pi^2} \cdot \frac{v^4}{g^2} \end{aligned}$
where we have used the half-angle formula $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$ and integrated by parts to evaluate $\int \cos 2\theta d\theta.$
Therefore, $c = \frac{45}{\pi^2}$
