Maximal area under projectile trajectory

Question

Problem:

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.

Then the maximal area under the trajectory is given by $k \cdot \frac{v^4}{g^2}$ for some constant $k.$ Find $k.$

My Work: I substituted the first equation for t into the second equation getting: $$y = \frac{\sin\theta x}{\cos\theta} - \frac{gx^2}{2v^2(\cos\theta)^2}.$$ How do I proceed?

Answer 1

Note that the projectile’s flying time in the air is $T=\frac{2v\sin\theta}g$. The area under the trajectory is

\begin{align} A =\int_0^T y(t)x’(t)dt=\int_0^T (vt \sin \theta - \frac{1}{2} gt^2)v\cos\theta dt=\frac{2v^4}{3g^2}\cos\theta\sin^3\theta \end{align}

Then, set $dA/d\theta=0$ to get the optimal angle $\theta =\frac\pi3$. Thus, the maximal area is

$$A_{max}=\frac{2v^4}{3g^2}\cos\frac\pi3\sin^3\frac\pi3=\frac{\sqrt3v^4}{8g^2} $$ and $k=\frac{\sqrt3}8$.

Answer 2

1. Find the range. Set $y=0$. One solution is $x=0$, divide by $x$, and solve for the other root.
2. Integrate to find the area. $$A=\int _0^{x_{max}} y dx$$
3. Take the derivative with respect to $\theta$ and set it to $0$: $$\frac{dA}{d\theta}=0$$
4. For the given $\theta$ calculate the area.