Area under the Trajectory of a projectile

Question

I may need some help with the following problem.
We have

$$ y(\omega)=a+\omega\cdot b - \frac{1}{2}\cdot\omega^2\cdot g $$ where $g$ is the gravity vector $g=\binom{0}{\lambda}$, $\lambda \approx 9,81\frac{m}{s^2}$, $a=\binom{0}{0}$ and $b=\dot y(0)=\binom{b_1}{b_2}$ with $b_1 \neq 0$.
The starting velocity $|\dot y(0)| $ is given.
The question is : Under which angle a projectile has to be launched to maximize the area under the parabolic path.
My Idea was to integrate $y$. Find the maximum by differetiating it (which would mean solving $y(\omega)=0$). And solving that for $b_1$. Is this correct?

Answer 1

If a projectile is fired from the origin at a velocity of $u \ \mathrm{ms}^{-1}$ at an angle of $\alpha^{\circ}$ with the horizontal.

The subsequent horizontal displacement at time $t$ is $s_x = ut\cos\alpha$ while the subsequent vertical displacement at time $t$ is $s_y=ut\sin\alpha - \tfrac{1}{2}gt^2$. The projectile returns to Earth when $s_y=0$. Solving $s_y=0$ gives $t=0$ and $t=2\tfrac{u}{g}\sin\alpha$. Clearly, $t=0$ is when the projectile was fired.

We are interested in the area under the curve with parametric equations $$s_x(t)=ut\cos\alpha \ \ \ \text{and} \ \ \ s_y(t)=ut\sin\alpha - \tfrac{1}{2}gt^2$$ between the times $t_1=0$ and $t_2=2\tfrac{u}{g}\sin\alpha$. The area under our parametric curve is $$A(\alpha) = \int_{t_1}^{t_2} s_y \ \frac{\mathrm{d}s_x}{\mathrm{d}t} \ \mathrm{d}t = \frac{2u^4}{3g^2}\cos\alpha\sin^3\alpha$$

To find the local maxima and local minima of $A$ we calculate the derivative with respect to $\alpha$: $$\frac{\mathrm{d}A}{\mathrm{d}\alpha} = \frac{2u^4}{g^2}(4\cos^2\alpha-1)\sin^2\alpha$$

The only stationary point with $0^{\circ} < \alpha < 90^{\circ}$ is $\alpha=60^{\circ}$. Calculating the second derivative: $$\left.\frac{\mathrm{d}^2A}{\mathrm{d}\alpha^2}\right|_{\alpha=\pi/3}=-\frac{u^4\sqrt{3}}{g^2} < 0$$ and so the projection angle $\alpha = 60^{\circ}$ maximises the area under the projectile.

EDIT

Below are two plots. Both have projection speed as $u=1 \ \mathrm{ms}^{-1}$ and gravity as $g=10 \ \mathrm{ms}^{-2}$. The red curve has $\alpha=60^{\circ}$ and the blue has $\alpha=45^{\circ}$.

Answer 2

I'm going to switch to using $t$ for the time variable. Hope you don't mind. :)

You'll get three equations:

$$x(t) = b_1t; \\ y(t) = b_2t - (1/2)\lambda t^2; \\ |v(0)| = \sqrt{b_1^2 + b_2^2}.$$

You can substitute the first into the second to get:

$$y(x) = \frac{b_2x}{b_1} - \frac{\lambda x^2}{2b_1^2}.$$

Now, also we know:

$$\tan \theta = \frac{b_2}{b_1},$$

and

$$b_1^2 = |v(0)|^2 \cos^2 \theta,$$

so

$$y(x) = x \tan \theta - \frac{\lambda x^2 \sec^2 \theta}{2 |v(0)|^2}.$$

The height is zero at $x=0$ and

$$x=\frac{2 \sin \theta \cos \theta |v(0)|^2}{\lambda} = \frac{|v(0)|^2 \sin 2\theta}{\lambda} ,$$

so these define our limits of integration. Integrating, we get

$$A(\theta) = \frac{2|v(0)|^4}{3\lambda^2}\sin^3 \theta \cos \theta = C \sin^3 \theta \cos \theta.$$

Finally, differentiating gives:

$$\frac{dA}{d\theta} = C \left(3 \sin^2 \theta \cos^2 \theta - \sin^4 \theta\right) = C \sin^2 \theta \left(3 \cos^2 \theta - \sin^2 \theta\right) = 0.$$

So $\sin \theta = n \pi$ (which we toss) or $\tan \theta = \pm \sqrt{3}$, which gives $\theta = \pi/3.$