Geometric Approach Explained

Okay, here’s how it goes. The trick is to make the right-angled triangle in the picture. The slope of the radial line is –h/a from rise over run, and that’s equal to the slope we found on the previous page, -1/(2a), which solves to give us h = 1/2.

Since the radius is 1, we get the following from Pythagoras:

h2 + a2 = 1
and thus:
(1/2)2 + a2 = 1
a2 = 1 - (1/2)2 = 3/4
Hence:

 
 

Finally let’s locate the centre of the circle. The parabola at x = a has height y = a2 so the centre of the circle has height

y = a2 + h = 3/4 + 1/2 = 5/4 = 1.25
and that determines the bottom of Quin’s bounce.