Bouncy Ball
Math brain teasers require computations to solve.
If you drop a bouncy ball from a distance of 9 feet from the floor, the ball will continue to bounce. Assume that each time it bounces back two thirds of the distance of the previous bounce. How far will the ball travel before it stops?
HintGeometric series...
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Answer
45 feet.
The distance can be calculated as 9 + (9)*(2/3)*(2) + (9)*(2/3)*(2/3)*(2)+..., with the 2 accounting for the upward and downward travel distance. Adding 9 to this will result in the formation of an infinite geometric series: 18*(2/3)^0 + 18*(2/3)^1 +... Factoring out the 18 will result in a series that is equal to (1/(1(2/3))), which equals 3. Multiplying by the 18 and then subtracting the nine used to create the series will result in 54  9 = 45.
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Comments
Unidus
Mar 10, 2002
 if the ball continues to bounce, like stated in the question, then the ball will never stop, and it will travel an infinite distance 
bobbrt
Jun 19, 2002
 I think that I can explain this mathematically: In the initial drop the ball will travel 9 ft. On the first bounce it will travel (9 ft) * (2/3) * 2 = 12 feet (the "2" is there because the ball travels the distance up and then on the way down again). On the second bounce it will travel (9 ft) * (2/3) * (2/3) * 2 = 8 ft. And on and on. Therefore the equation to be used is D = 9 + sum of [9* 2 * (2/3)^n], where D is the total distance traveled and n is the number of bounces. If you plug this into the engineer's best friend "MathCad", you get the answer of 45 feet. The reason the ball doesn't travel an infinite distance is that this is an exponential decay  as the number of bounces gets higher, the distance is so small that it doesn't increase the total distance traveled by a significant amount. 
jimbo
Dec 17, 2002
 Answer is 45 feet. It is a geometric progression. If the ball takes 1 second to complete the first bounce and two thirds of the time for each successive bounce, how long will it take before it stops bouncing and how many bounces will it have done? 
Rowsdower
Dec 23, 2003
 Wow, bobbrt and Jimbo are the guys to talk to for help with math! 
canu
Jul 13, 2004
 I wish jimbo had stated under what experimental conditions he would observe his bouncing ball.
It is not on the surface of the Earth, where the first bounce (6 feet up and 6 feet down) would take about 1.22 seconds.
It is not on any other planet, nor in any situation where the accelaration of gravity is constant, because there each successive bounce would take SQUARE ROOT OF 2/3 the time of the previous one (thus on Earth, the 2nd bounce would take about 1 second).
It must have been in a rocket undergoing a constantly increasing acceleration. I will let anybody else describe the specifications of such a rocket.
I don't have MathCad but I remember something of the freshman high school physics of 60 years ago.

Atropus
Feb 09, 2005
 If it is exactly 2/3 decay then it keeps going infinitely doesn't it? or until the movement is so infinitesimal that it gets molecular ^_^ bwahah 
Sane
Mar 23, 2005
 Troppy, it keeps thirding though, so no matter how many more bounces you make it do at such a small height, it technically would never get to 51.
Like 1/1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 ...
...never gets to 2/1. 
darthforman
May 27, 2005
 Who cares.

musicmaker21113
Jan 09, 2006
 I agree with Sane. Technically, the way this teaser is phrased, this ball will never stop bouncing....as it always bounces back up 2/3rds the distance it dropped. It will keep getting closer and closer to 45 feet, but will never actually get that far. 
mr_brainiac
Jan 10, 2006
 I need to reboot my brain. I'm suffering mental brownouts again. I had my choo choo on the right track but somehow I got derailed. 
SPUTNIK2
Jan 13, 2006
 very clever! 
keveffect1
Feb 21, 2006
 Nice teaser to remind me of how use geometrics 
jntrcs
Mar 18, 2006
 I think the answer is infinite. think of it this way: It gets 2/3 size so it is growing continually and even though it would get so small of a bounce nothing could determine it. (Although odly like pi) but it keeps going and getting larger. I'm no scientist but that is how i see it. 
albuquer
May 12, 2006
 i agree that the ball will never stop bouncing because of the wording of the question, it travels an infinite distance, but yes if you talking exponential decay a limits, there would be a mathematical answer you can come up with, but the way the answer was worded it would never stop "traveling" if it travels one millionth of a foot one million times, thats another foot 
Smudge
May 25, 2006
 It would never stop, as is the wording of the question.
Nor would it ever reach the stated answer of 45 feet.
When you all say that it would go an infinite distance, you're wrong.
Yes, it would bounce an infinite number of times, but the distance for each bounce is so small that it would just keep adding another decimal place to the distance travelled, such that it would be 44.99999999999999999999... for a hell of a long time (which is infinity, i guess). 
Smudge
May 25, 2006
 it's like if you add 1 + 1/10 + 1/100 + 1/1000 + 1/10000 + 1/100000 + 1/1000000 + ...
after one iteration  1
after two iterations  1.1
after three iterations  1.11
after four iterations  1.111
after five iterations  1.1111
after six iterations  1.11111
after...... 
babygirl8195
Jun 08, 2006
 what? 
udoboy
Jun 19, 2006
 Dear "Sane"
1/1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 ...
Is already greater than 2/1 when you add the 1/4 term.
It's the summation of (1/2ⁿ); n ≥ 0 that doesn't reach 2. 
xdbtcp
Jun 21, 2006
 used excel  quit at 44.99972.... 
GebbieRose
Aug 11, 2006
 Close enough! Nice teaser. 
flowerz2010
Jan 30, 2007
 Wouldn't the ball never get to the floor, but very close because 2/3 of 1 is 2/3 and to get 2/3 of that you would have to multiply it by 2/3. Therefore, the ball may get very close to landing on the floor, but would still be bouncing. 
flowerz2010
Jan 30, 2007
 Wouldn't the ball never get to the floor, but very close because 2/3 of 1 is 2/3 and to get 2/3 of that you would have to multiply it by 2/3. Therefore, the ball may get very close to landing on the floor, but would still be bouncing. 
Jerrythellama
Feb 04, 2007
 ah, good old calculus 
bgil7604
Feb 21, 2007
 I didn't think the instructions were written well eneough. I thought it was a trick question, making it 0, but still really cool. 
musa_karolia
Apr 11, 2007
 the answer will tend to 45 without reaching it .... move on. ty 
javaguru
Jan 28, 2009
 After 198 bounces, the distance the ball would travel on the 199th bounce is less than the Planck length (1.616 x 10^35 meters = 5.30 x 10^35 feet). At that point, no smaller unit of movement is possible and the ball is either at at rest, or more likely, is oscillating between the two positions. Either way, the precision of the measurement of the position of the ball is far beyond the limits of what the Heisenberg uncertainty principle allows.
Disregarding the effect of quantized space, which would mean that the distance the ball travels is limited to multiples of the length of the quanta, the answer is then
45 +/ 5.3 x 10^35 feet
I just rounded this to 45 feet given that the effects of quantized space make the error unmeasurable.

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