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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?

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It is very similiar to my question. Look at that physics.stackexchange.com/questions/28863/… –  Mathlover Oct 19 '12 at 12:34

1 Answer 1

up vote 2 down vote accepted

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.

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The geometric sum of horizontal distance after infinite bounces would be finite (e.g if each bounce is halved, the sum is 2m in total after infinite bounces). But you assume horizontal velocity doesn't change. Will the ball travel forever or just 2m? –  Mew Oct 18 '12 at 14:48
    
Good question: that's Zeno's paradox isn't it? Assuming the exponential relation holds exactly, the ball will start bouncing infinitely quickly as the horizontal distance approaches whatever 1 + 1/4 + 1/16 + etc is. It will therefore lose energy infinitely quickly and stop bouncing. beyond this point the ball will roll without bouncing. –  John Rennie Oct 18 '12 at 15:01
    
Thanks, nice explanation. –  Mew Oct 18 '12 at 15:03

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