How do balls bounce on slopes?

It is easier to view the situation, not on a slope, but on a horizontal surface. We give gravity a horizontal and vertical component.

Since the vertical component of gravity is constant, the ball must bounce to the same height, $h$ , above the slope each time, which means that the time between consecutive collisions is constant.

Horizontally, the distance between collisions, $s$ , can be given as $s=ut+\frac{1}{2}at^2$ . Since the horizontal acceleration and the time taken are both constant, we can rewrite this equation as $s=Cu+D$ (remember that $u$ is the horizontal component of velocity).

Now, because there is constant horizontal acceleration, the horizontal velocity $u$ increases by the same amount with each bounce. We could, for the sake of things, write this as $u_n=K+u_{n-1}$ , where $u_n$ is the horizontal velocity after the $n^{th}$ collision.

What does this tell us? Well since the distance between each bounce is $Cu+D$ , and $u$ increases by a constant amount between collisions, there must be a constant difference between the distances of each bounce.

And thus the resultant progression is $\boxed{\text{arithmetic}}$ .