The arc of high jumping bends towards ... 2?

In the high jump, an athlete’s entire body must clear the bar. However, not every part of their body has to clear the bar at the same time. As a result, athletes arc their bodies over the bar, so that only a fraction of their mass is above the bar at any given time. In fact, athletes can theoretically clear the bar despite their center of mass remaining below the bar throughout the jump.

Let’s model the athlete’s jump as an arc of angle $2\phi$ that is centered over the bar, as shown below. For simplicity, assume that their mass is uniformly distributed across the length of their body. The dot in the diagram represents the athlete’s center of mass. Let $a$ represent the vertical distance between the athlete’s center of mass and their lowest points (presumably their outstretched fingers and toes), and let $b$ represent the vertical distance between the athlete’s center of mass and their highest point (presumably their waist).

If $\phi = \frac{\pi}{2}$ then the arc would be a complete semicircle. In that case, what is the ratio $a/b$?

As the angle $\phi$ gets very, very small (i.e., in the limit as $\phi$ goes to zero), what value does the ratio $a/b$ approach?

Let's solve the generic case for $\phi$ so that we can then do the two cases, both when $\phi = \frac{\pi}{2}$ and when $\phi \downarrow 0.$ Since we'll be taking ratios of everything anyway, let's just go ahead and assume that the circle on which the high jumper's arc is lying in the figure has radius $1$ and is centered at the origin. Then, with some trigonometric know-how, we see that the lowest point of the arc will be sitting at a height of $\cos \phi.$ So we must have $a + b = 1 - \cos \phi.$ Additionally we see that the jumper's body has a length of $2\phi,$ the length of this arc. So to find $a,$ we need only take the average of the jumper's height along this arc by taking the integral

$$a + \cos \phi = \frac{1}{2\phi} \int_{\frac{\pi}{2} - \phi}^{\frac{\pi}{2} + \phi} \sin t \,dt = \frac{1}{2\phi} \left( -\cos (\frac{\pi}{2} + \phi) + \cos (\frac{\pi}{2} - \phi) \right) = \frac{\cos (\frac{\pi}{2} - \phi) }{\phi} = \frac{\sin \phi}{\phi}.$$ That is, $$a = \frac{\sin \phi}{\phi} - \cos \phi$$ and $$b = 1 - \cos \phi - a = 1 - \frac{\sin \phi}{\phi}.$$ Therefore for any value of \phi, the ratio of $a / b$ is $$r(\phi) = \frac{\frac{\sin \phi}{\phi} - \cos \phi}{1 - \frac{\sin \phi}{\phi}} = \frac{ \sin \phi - \phi \cos \phi }{\phi - \sin \phi}.$$

So in particular, if $\phi = \frac{\pi}{2}$ and the jumper's body makes a complete semicircle then the ratio $a/b$ is

$$r(\frac{\pi}{2}) = \frac{\sin \frac{\pi}{2} - \frac{\pi}{2} \cos \frac{\pi}{2} }{\frac{\pi}{2} - \sin \frac{\pi}{2}} = \frac{1}{\frac{\pi}{2} - 1} = \frac{2}{\pi - 2} \approx 1.75193839388....$$

On the other hand, if we use some rough Taylor approximations around $\phi \approx 0$ we see that we have

$$r(\phi) = \frac{\left(\phi - \frac{\phi^3}{6} + O(\phi^5)\right) - \phi \left( 1 - \frac{\phi^2}{2} + O(\phi^4)\right)}{\phi - \left(\phi - \frac{\phi^3}{6} + O(\phi^5)\right)} = \frac{ \frac{\phi^3}{3} + O(\phi^5) }{\frac{\phi^3}{6} + O(\phi^2)} = 2 \frac{1 + O(\phi^2)}{1 + O(\phi^2)}.$$ So, as $\phi \downarrow 0,$ we have $r(\phi) \to 2.$