How High Can You Jump?

This Week's Fiddler

Because we only care about the ratio of $a$ to $b$, we can work with a unit circle, as any radius dependency will cancel out. Let's map the scenario to the following (crudely drawn) coordinate system, where the high jumper's body is an arc of the unit circle and $(0, y)$ is their center of mass.

In this case, $a=y$ and $b=1-y$, so we wish to find $\frac{y}{1-y}$.

The center of mass would be the average of all constituent points on the arc, and we want to find the $y$–value of said center. We can achieve this with an integral using polar coordinates, finding for any given angle $\theta$ the $y$–value at that angle ($\sin\theta$). We use polar coodinates to ensure that the "density" of points remains constant throughout the integration. We want to also divide by the total arc length or "weight": $\pi$ (the semiperimeter of a unit circle).

This gives us the expression:

$$\begin{align*} y&=\frac{1}{\pi}\int_0^\pi \sin\theta\ \textrm{d}\theta \\[1em] &=\frac{1}{\pi}\biggl[-\cos\theta\biggr]_0^\pi \\[1em] &=\frac{1}{\pi}\bigl(\cos0-\cos\pi\bigr) \\[1em] &=\frac{2}{\pi} \end{align*}$$

So $a=\frac{2}{\pi}$ and $b=1-\frac{2}{\pi}$, making our answer:

$$\begin{align*} \frac{a}{b}&=\frac{\frac{2}{\pi}}{1-\frac{2}{\pi}} \\[1em] &=\frac{2}{\pi-2} \end{align*}$$

Answer: 2/(pi - 2), or about 1.75

Extra Credit

Let's consider what happens when the body is an arc of $2\phi$ instead of just a semicircle.

Let's use the same coordinate system as the previous part. Note that the body's lowest point now has a $y$–value of $\cos\phi$ instead of $0$.

For this shape, we can use a similar integration technique over $\theta$, this time going from $\frac{\pi}{2}-\phi$ to $\frac{\pi}{2}+\phi$ instead of from $0$ to $\pi$. This means our arc length is also no longer $\pi$; it's $2\phi$. So $y$, the $y$–value of our center of mass, is expressed as follows:

$$\begin{align*} y&=\frac{1}{2\phi}\int_{\frac{\pi}{2}-\phi}^{\frac{\pi}{2}+\phi} \sin\theta\ \textrm{d}\theta \\[1em] &=\frac{1}{2\phi}\biggl[-\cos\theta\biggr]_{\frac{\pi}{2}-\phi}^{\frac{\pi}{2}+\phi} \\[1em] &=\frac{1}{2\phi}\biggl(\cos{\Bigl(\frac{\pi}{2}-\phi\Bigr)}-\cos{\Bigl(\frac{\pi}{2}+\phi\Bigl)}\biggr) \\[1em] &=\frac{1}{2\phi}\Bigl(\sin\phi-\sin{(-\phi)}\Bigr) \\[1em] &=\frac{1}{2\phi}\Bigl(2\sin\phi\Bigr) \\[1em] &=\frac{\sin\phi}{\phi} \end{align*}$$

This corroborates our previous part's finding of $y=\frac{2}{\pi}$ for $\phi=\frac{\pi}{2}$. The expression $y=\frac{\sin\phi}{\phi}$ also famously approaches $1$ as $\phi\to0$, which checks out intuitively: the center of mass should approach $y=1$ as the body moves further and further up the coordinate system. But what about our $\frac{a}{b}$ ratio?

Because the tips of the body now start at a $y$–value of $\cos\phi$ instead of $0$, $a$ needs to be adjusted to $y-\cos\phi$. Note that $b$ is still $1-y$ because the $y$–value of the highest point is still $1$. So we wish to find the ratio of $\frac{y-\cos\phi}{1-y}$ as $\phi\to0$:

$$\begin{align*} \lim_{\phi\to0}\frac{a}{b}&=\lim_{\phi\to0}\frac{y-\cos\phi}{1-y} \\[1em] &=\lim_{\phi\to0}\frac{\frac{\sin\phi}{\phi}-\cos\phi}{1-\frac{\sin\phi}{\phi}} \\[1em] &=\lim_{\phi\to0}\frac{\sin\phi-\phi\cos\phi}{\phi-\sin\phi} \\[1em] &=\frac{0}{0} \end{align*}$$

We can use L'Hôpital's Rule here!

$$\begin{align*} \lim_{\phi\to0}\frac{a}{b}&=\lim_{\phi\to0}\frac{\frac{\textrm{d}}{\textrm{d}\phi}(\sin\phi-\phi\cos\phi)}{\frac{\textrm{d}}{\textrm{d}\phi}(\phi-\sin\phi)} \\[1em] &=\lim_{\phi\to0}\frac{\cos\phi+\phi\sin\phi-\cos\phi}{1-\cos\phi} \\[1em] &=\lim_{\phi\to0}\frac{\phi\sin\phi}{1-\cos\phi} \\[1em] &=\frac{0}{0} \end{align*}$$

And again...

$$\begin{align*} \lim_{\phi\to0}\frac{a}{b}&=\lim_{\phi\to0}\frac{\frac{\textrm{d}}{\textrm{d}\phi}(\phi\sin\phi)}{\frac{\textrm{d}}{\textrm{d}\phi}(1-\cos\phi)} \\[1em] &=\lim_{\phi\to0}\frac{\phi\cos\phi+\sin\phi}{\sin\phi} \\[1em] &=\frac{0}{0} \end{align*}$$

And again!

$$\begin{align*} \lim_{\phi\to0}\frac{a}{b}&=\lim_{\phi\to0}\frac{\frac{\textrm{d}}{\textrm{d}\phi}(\phi\cos\phi+\sin\phi)}{\frac{\textrm{d}}{\textrm{d}\phi}(\sin\phi)} \\[1em] &=\lim_{\phi\to0}\frac{-\phi\sin\phi+\cos\phi+\cos\phi}{\cos\phi} \\[1em] &=\lim_{\phi\to0}\frac{2\cos\phi-\phi\sin\phi}{\cos\phi} \\[1em] &=2 \end{align*}$$

Answer: 2