### Brain Teasers

# Bouncy Ball

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?

### Hint

Geometric series...### 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.

Hide Hint Show Hint Hide Answer Show Answer

## What Next?

**Solve a Similar Brain Teaser...**

Or, get a random brain teaser.

If you become a registered user you can vote on this brain teaser, keep track of

which ones you have seen, and even make your own.

## Comments

if the ball continues to bounce, like stated in the question, then the ball will never stop, and it will travel an infinite distance

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.

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?

Wow, bobbrt and Jimbo are the guys to talk to for help with math!

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.

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.

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

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.

Like 1/1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 ...

...never gets to 2/1.

Who cares.

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.

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.

very clever!

Nice teaser to remind me of how use geometrics

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.

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

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).

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).

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......

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......

what?

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 (1/2â¿); n â‰¥ 0 that doesn't reach 2.

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 (1/2â¿); n â‰¥ 0 that doesn't reach 2.

used excel - quit at 44.99972....

Close enough! Nice teaser.

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.

ah, good old calculus

I didn't think the instructions were written well eneough. I thought it was a trick question, making it 0, but still really cool.

the answer will tend to 45 without reaching it .... move on. ty

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.

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.

## Follow Braingle!