If a ball is simply dropped, each time a ball bounces, it's height decreases in what appears to be an exponential rate.
Let's suppose that the ball is thrown horizontally instead of being simply dropped. How does the horizontal distance travelled change after each bounce?
Context behind question: I read a question involving a ball that travels horizontally 1m during the first bounce, 0.5m during the second bounce, 0.25 during the third bounce etc. I was wondering if this model is physically valid?
Best Answer
If a bounce has a maximum height $h$ then the time taken for the ball to leave the ground, reach $h$ and fall back is simply $t = 2\sqrt{2h/g}$ so if the horizontal speed is constant at $v$ the horizontal distance travelled in the bounce is $s = 2v\sqrt{2h/g}$.
In your example the distance travelled appears to halve with each bounce i.e. $s_{n+1}/s_n = 1/2$. Since $s \propto \sqrt{h}$ we get:
$$ \frac{s_{n+1}}{s_n} = \sqrt{\frac{h_{n+1}}{h_n}} = \frac{1}{2}$$
so $h_{n+1}/h_n = 1/4$. This seems physically reasonable and it is an exponential decay of bounce height.
NB this all assumes the horizontal velocity doesn't change at the bounce.