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### Course: Algebra (all content)>Unit 18

Lesson 8: Infinite geometric series applications

# Infinite geometric series word problem: bouncing ball

Watch Sal determine the total vertical distance a bouncing ball moves using an infinite geometric series. Created by Sal Khan.

## Want to join the conversation?

• Would anyone mind explaining the -10 to me, please?
• The ball bounces half as high each time right?
First he shows that the sum of the distance traveled is:
10m - the original height of the ball, plus
10(1/2)m + 10(1/2)m = (10+10)(1/2) = 20(1/2), which is the ball bouncing back 1/2 the original height, 5m and falling back down, plus
10(1/2)(1/2) + 10(1/2)(1/2) = 20(1/2)^2, which is the ball bouncing half as high again as the previous bounce (which was 10(1/2)), plus and on and on.
So it looks like the sequence could be written as S=20(1/2)^n as n goes from 0 to infinity right?
EXCEPT the ball only FELL the first time, it did not BOUNCE up and down like all the successive bounces do, so it only traveled the 10m ONCE. So if we want to write the series as S=20(1/2)^n as n goes from 0 to infinity, that series says the ball bounced up 10m and back down 10m, but that is not what happened, it only fell 10m, therefore we need to REMOVE the extra 10m that the term 20(1/2)^0 adds to the sum.
Did that help?
Keep Studying!
• If I start with the `k=1`th term, `20(1/2)^1`, I get `20/(1-(1/2))`for `a/(1-r)`.
Now, the first term is `10`.
So I get `10/(1-(1/2)) = 10/(1/2) = 20`.
If I then add the term for `k=0`, which is `10`, I get `20+10=30`, which is the same answer Sal got by substituting `-10+20` for the `10`.
Is his method preferable to mine, or doesn't it make a difference? Why or why not?
• I think I have found an elegant general solution to this problem. If the initial height is represented by h and the fraction of the height it bounces is (1/b), it can be shown that the total vertical distance D is:
D = h * (b + 1) / (b – 1)
The derivation is as follows:
D = h + Summation of k=0 to infinity[ 2 * (h/b) * (1/b)^k ]
D = h + (2h/b) / (1 – (1/b))
D = h + (2h/b) / ((b – 1)/b)
D = h + (2h / (b – 1))
D = ((bh – h)/(b – 1)) + (2h/(b – 1))
D = (bh – h + 2h) / (b – 1)
D = (bh + h) / (b – 1)
Thus:
D = h * (b + 1) / (b – 1)
• A bouncing ball on the floor would stop bouncing after a certain finite amount of bounces right? So why do we use the infinite formula?
• Because a more realistic mathematical model of a bouncing ball is much too advanced for your present level of mathematics. So, this is the best that can be done at this level of math.

And, the final answer isn't too far off what would really happen because in this model after the first several bounces the ball would be moving so little that all those infinitely many bounces it describes would not really add much of anything to the total distance traveled.
• Couldn't we just find the vertical distance in one direction with a basic process and multiply the result times two?
• Actually, that does not work. Notice that the first time the ball is introduced in Sal's diagram, it is dropped from 10 ft. But it doesn't go up to 10 ft again after it was dropped, nor did it go up to 10 ft before it was dropped as a result of a bounce. So we have to either sum two series, one starting at 10ft and another starting at 5 ft, with common ratio 1/2, or we have to do Sal's method.
If you were to multiply by 2, and only have one series starting at 10 ft, then you would have to subtract 10 ft at the end to account for over-counting.
• Just had a dynamics exam, and my professor wanted us to find the time traveled for a .5 lb ball that was released from a height of 10 meters and a coefficient of restitution of .9. He said after the test that we had to represent that answer as an infinite series. How would that be accomplished?
• At , why does Sal write -10, is this because the ball is going down to the ground according to physics or he just make typo?
• It depends how you want to formulate your series.
In the video Sal starts the series at n = 0.
What is 20 * (1/2)^0?
It is 20.
However, the first bounce is only 10m, not 20m!
Therefore, we need to deduct 10.
In other words, after the ball has been dropped (i.e., the 0th bounce) it has travelled a distance of:
-10 + 20(1/2)^0 = 10m
• I used a different method and got the same result:

10+(Σ 5*(1/2)^k)*2=30, bounds from k=0 to ∞

This way is also correct, right?
• Yes I'm pretty sure this works. I did it a similar way except instead of multiplying the whole series by 2 at the end I just did the sum of 10(1/2)^k instead of 5(1/2)^k. I got the same answer.
• why does he add a negative ten??
(1 vote)
• Because he replaced an existing positive ten with a positive twenty plus a negative ten.
10 = 20 - 10 = -10 + 20

Also note that 20 = 20 * 1 = 20 * (1/2)^0
• At , he wrote it as to the power 6 instead of 0. Am I mistaken?
• Supposed to be 0, just added a little tail to the 0 accidentally I assume.
(1 vote)
a ball dropped from 10 metres rebounds 3/4 of the distance it fell. how can i find the total vertical distance that the ball travels until the moment it hits the floor for the tenth time