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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.

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  • male robot johnny style avatar for user dennis
    Would anyone mind explaining the -10 to me, please?
    (23 votes)
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    • leaf blue style avatar for user Stefen
      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)
  • male robot hal style avatar for user ledaneps
    If I start with the k=1th 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)
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    • male robot hal style avatar for user George Arrington
      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)
  • leaf green style avatar for user Salcott
    Couldn't we just find the vertical distance in one direction with a basic process and multiply the result times two?
    (4 votes)
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    • marcimus pink style avatar for user Alex Tran
      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)
  • starky tree style avatar for user sixhundredandsixtysix
    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)
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    • piceratops ultimate style avatar for user Just Keith
      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)
  • spunky sam blue style avatar for user Tyler Huttenlocher
    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)
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  • aqualine tree style avatar for user Nevergrowingup16
    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?
    (2 votes)
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    • aqualine ultimate style avatar for user David
      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)
  • mr pink red style avatar for user Ariana
    why does he add a negative ten??
    (1 vote)
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  • hopper cool style avatar for user YYCheng.khan
    At , he wrote it as to the power 6 instead of 0. Am I mistaken?
    (3 votes)
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  • blobby green style avatar for user Carly Cody
    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)
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    • aqualine seed style avatar for user John Waggoner
      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)
  • blobby green style avatar for user Praise Melchizedek
    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)
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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.