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## Integral Calculus

### Unit 5: Lesson 2

Infinite geometric series- Worked example: convergent geometric series
- Worked example: divergent geometric series
- Infinite geometric series
- Infinite geometric series word problem: bouncing ball
- Infinite geometric series word problem: repeating decimal
- Proof of infinite geometric series formula
- Convergent & divergent geometric series (with manipulation)

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# Infinite geometric series word problem: bouncing ball

AP.CALC:

LIM‑7 (EU)

, LIM‑7.A (LO)

, LIM‑7.A.3 (EK)

, LIM‑7.A.4 (EK)

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?(23 votes)
- 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!(50 votes)

- 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?

Thanks in advance for your insight.(8 votes)- 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)(19 votes)

- Couldn't we just find the vertical distance in one direction with a basic process and multiply the result times two?(4 votes)
- 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.(6 votes)

- 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?(3 votes)
- 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.(6 votes)

- 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?(2 votes)
- At3:20, why does Sal write -10, is this because the ball is going down to the ground according to physics or he just make typo?(2 votes)
- 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(4 votes)

- 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(5 votes)

- At3:45, he wrote it as to the power 6 instead of 0. Am I mistaken?(3 votes)
- Supposed to be 0, just added a little tail to the 0 accidentally I assume.(1 vote)

- please help!!

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(2 votes)- for any n (number of repetitions), [ k = n - 1 ] to get n terms

for a geometric series, sum: [a(r^k) from k=0 to k=9]

becomes

sum: [10({3/4}^k) from k=0 to k=9]

and for geometric series the [sum] = [ a(1-r^n)/(1-r) ]

this gives us 10(1 - {3/4}^10) / (1 - 3/4) = 4947635/131072

or approximately 37.74746

Assuming I did all that without an error.(2 votes)

- Can someone answer this please? If an infinite geometric series has no sum (since r is equal or greater than 1), what does that mean given a situation?(1 vote)
- Well, if there's no sum, then the series will add up to positive or negative infinity, or oscillate. (I accidentally forgot the oscillation part, credit to CPT Winslow.)(2 votes)

## Video transcript

Let's say that we have
a ball that we dropped from a height of 10
meters, and every time it bounces it goes half as
high as the previous bounce. So for example, you
drop it from 10 meters. The next time its peak height
is going to be at 5 meters. So the next time around,
on the next bounce, let me draw in that
same orange color. And the next bounce the ball
is going to go 5 meters. This distance right over
here is going to be 5 meters. And then the bounce after that
is going to be half as high. So it's going to go
2 and 1/2 meters. And it's just going
to keep doing that. So it's going to go 2 and
1/2 meters right over here. And what I want to think
about in this video is what is the total vertical
distance that the ball travels? So let's think about
that a little bit. So it's first going to travel
10 meters straight down. So it's going to travel
10 meters just like that, and then it's going to travel
half of 10 meters twice. It's going to go up 5
meters, up half of 10 meters, and then down half of 10 meters. Let me put it this way. So each of these is
going to be 10 meters. Actually, I don't have
to write the units here. Let me take the
units out of the way. Let me write that clear. So the first bounce,
once again, it goes straight down 10 meters. Then on the next bounce it's
going to go up 10 times 1/2. And then it's going to
go down 10 times 1/2. Notice we just care about
the total vertical distance. We don't care about
the direction. So it's going to go up 10
times 1/2, up 5 meters, and then it's going
to go down 5 meters. So it's going to travel a total
vertical distance of 10 meters, 5 up and 5 down. Now what about on this jump, or
on this bounce, I should say. Well here it's going to go
half as far as it went there. So it's going to go 10 times 1/2
squared up, and then 10 times 1/2 squared down. And I think you
see a pattern here. This looks an awful lot
like a geometric series, an infinite geometric series. It's going to just keep on going
like that forever and ever. So let's try to clean this
up a little bit so it looks a little bit more like a
traditional geometric series. So if we were to simplify
this a little bit we could rewrite
this as 10 plus 20. 20 times 1/2 to the first power,
plus 10 1/2 times 1/2 squared plus 10 times 1/2 squared
is going to be 20 times 1/2 squared, and we'll just
keep on going on and on. So this would be a
little bit clearer if this were a 20
right over here. But we could do that. We could write 10 as
negative 10 plus 20, and then we have plus all of
this stuff right over here. Let me just copy and paste that. So plus all of this
right over here. And we can even
write this first. We can even write this 20
right over here is 20 times 1/2 to the 0 power
plus all of this. So now it very
clearly looks like an infinite geometric series. We can write our entire
sum, and maybe I'll write it up here since I don't
want to lose the diagram. We could write it
as negative 10. That's that negative
10 right over here. Plus the sum from
k is equal to 0 to infinity of 20 times our
common ratio to the k-th power. So what's this going to be? What's this going
to turn out to be? Well we've already
derived in multiple videos already here that the sum of an
infinite geometric series, so the sum from k equals 0
to infinity of a times r to the k is equal
to a over 1 minus r. So we just apply
that right over here. This business right
over here is going to be equal to 20
over 1 minus 1/2, which is the same
thing as 20 over 1/2, which is the same thing
as 20 times 2, or 40. So what's the total vertical
distance that our ball travels? It's going to be negative
10 plus 40, which is equal to 30 meters. Our total vertical distance that
the ball travels is 30 meters.