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bayley86 · Jan 7, 2010, 02:11 PM

A ball is dropped from a height of 80m. It bounces to a height equivalent to 0.8 of the original height. Calculate the total distance the ball will travel when it eventually stays at rest. (you will have to think about infinite series techniques)

Can anyone help me with this I have no idea about infinite series.

Thanks

Andy

galactus · Jan 7, 2010, 02:32 PM

Originally Posted by bayley86

A ball is dropped from a height of 80m. it bounces to a height equivalent to 0.8 of the original height. calculate the total distance the ball will travel when it eventually stays at rest. (you will have to think about infinite series techniques)

can anyone help me with this i have no idea about infinite series.

thanks

Andy

Picture what is going on. It is a geometric series. Theoretically, the ball bounces forever, but we know that is not really how it is.

But, we drop the ball an initial 80 meters. Then it comes back up 80% of that, then back down the same distance, then back up 80% of that, then back down, and on and on.

Downward series $80+.8(80)+.8(.8(80))+......$

$80+64+51.2+40.96+.....................$

Upward series is the same as before because it falls as far as it goes up. Right?

$64+51.2+40.96+.............$

The distance the ball travels is found by adding this infinte series.

$S=80+2(64+51.2+40.96+..................)$

$S=80+2\left(64+64(.80)+64(.80)^{2}+64(.80)^{3}+... .\right)$

$S=80+128(1+.80+(.80)^{2}+(.80)^{3}+.....\right)$

$S=80+128\sum_{k=0}^{\infty}(\frac{4}{5})^{k}$

Now, we can't really add up an infinite amount of numbers, so we use the general formula for the geometric series.

$S=\frac{a_{1}}{1-r}$

Where $a_{1}$ is the first term (64) and r is the common ratio (4/5=.80).

The common ratio is .8 or 4/5

$S=80+2\left(\frac{64}{1-\frac{4}{5}} \right)$

$=80+128(\frac{1}{1-\frac{4}{5}})=720$ .

The ball travels 720 meters.

See there? I hope this helps for future problems. Keep this as a tutorial.