Pendulum Projectile

Let's start off with the equations of motion for our ball after it is released from the string. Since we are calculating horizontal distance by the time the ball returns to its launch height, we can let the origin coincide with the point where the ball becomes detached from the string. We then have:

$x(t) \ = \ x_0 \ + \ v \cos \theta \ t$ $y(t) \ = \ y_0 \ + \ v \sin \theta \ t \ - \ \frac{1}{2}gt^2$

Rearranging the first equation for $t$ and plugging it into the second equation, we get:

$t \ = \ \frac{x \ - \ x_0}{v \cos \theta}$ $\Rightarrow \ y(x) \ = y_0 \ + \ (x \ - \ x_0) \tan \theta \ - \ \frac{g(x \ - \ x_0)^2}{2v^2 \cos^2 \theta}$

Now, to find $v$ , which is the initial velocity of the ball when it is launched, we simply note the ball begin in the horizontal and swung down. The height of the ball when it is located at angle $\theta$ will be (from basic trigonometry), $L \ - \ L \cos \theta$ . This means that:

$E_0 \ = \ mgL$ $E_{f} \ = \ mgL(1 - \cos \theta) \ + \ \frac{1}{2} mv^2$ $E_0 \ = \ E_f \ \Rightarrow \ mgL \ = \ mgL(1 - \cos \theta) \ + \ \frac{1}{2} mv^2 \ \Rightarrow \ v \ = \ \sqrt{2gL \cos \theta}$

Plugging this into the equation for $y(x)$ , we get:

$y(x) \ = y_0 \ + \ (x \ - \ x_0) \tan \theta \ - \ \frac{(x \ - \ x_0)^2}{4L \cos^3 \theta}$

We are interested in the total vertical range of the ball, therefore we set $y(x) \ = \ y_0$ , and get:

$(x \ - \ x_0) \tan \theta \ - \ \frac{(x \ - \ x_0)^2}{4L \cos^3 \theta} \ = \ 0$

Let's declutter this expression, we are interested in the change in $x$ after the projectile is launched, therefore we can omit $x_0$ . Rearranging for a function of $x$ with respect to $\theta$ :

$x \tan \theta \ = \ \frac{x^2}{4L \cos^3 \theta} \ \Rightarrow \ \tan \theta \ = \ \frac{x}{4L \cos^3 \theta} \ \Rightarrow \ x(\theta) \ = \ 4L \tan \theta \cos^3 \theta$

We can find the maximum of this function by taking the derivative and setting $x'(\theta) \ = \ 0$ :

$\frac{\text{d}}{\text{d}\theta} \ x(\theta) \ = \ \frac{\text{d}}{\text{d}}\theta \ 4L \tan \theta \cos^3 \theta \ = \ 4L(cos^3 \theta \ - \ 2 \cos \theta \sin^2 \theta)$ $\Rightarrow \ 4L(cos^3 \theta \ - \ 2 \cos \theta \sin^2 \theta) \ = \ 0 \ \Rightarrow \ cos^3 \theta \ = \ 2 \cos \theta \sin^2 \theta \ \Rightarrow \ \tan \theta \ = \ \frac{1}{\sqrt{2}} \ \Rightarrow \ \theta \ = \ \tan^{-1} \ \frac{1}{\sqrt{2}}$

So this means that our answer is $x \ = \ \frac{1}{\sqrt{2}} \ \approx \ 0.707$ .