Geometric Sequence: Find 4th Bounce | Get Steps

I am sorry, but I am just a computer program and do not have access to previous posts. Could you please clarify what you are asking for? In summary, the conversation was discussing a problem about a ball dropped from a height of 3 feet with an elasticity that makes it bounce up one-third the distance it has fallen. The question asked for the total distance traveled by the ball at the instant it hits the ground the fourth time. The conversation also included discussions about visualizing the problem, drawing graphs, and finding a formula for the total distance traveled at the nth bounce. After multiple attempts, the correct answer was found to be 9 feet minus the sum of a geometric series with a common ratio of 1/3.
  • #1
js14
43
0
A ball is dropped from a height of 3 ft. The elasticity of the ball is such that it always bounces up one-third the distance it has fallen.
(a) Find the total distance the ball has traveled at the instant it hits the ground the fourth time. (Enter your answer as an improper fraction.) Can someone show me the steps oon how to do this?
 
Physics news on Phys.org
  • #2
You could start by drawing a graph of the ball's altitude vs time (does not need to be exact with regard to time) to better visualize what is going on.
 
Last edited:
  • #3
js14 said:
A ball is dropped from a height of 3 ft. The elasticity of the ball is such that it always bounces up one-third the distance it has fallen.
(a) Find the total distance the ball has traveled at the instant it hits the ground the fourth time. (Enter your answer as an improper fraction.) Can someone show me the steps oon how to do this?

How far would the ball drop the first time?
How far would it come up?
How far would it fall the second time?
How far would it come up?
How far would it fall the third time?
How far would it come up?
How far would it fall the fourth time?

What is the sum of all these distances?
 
  • #4
so am I subtracting every time the ball bounces?
 
  • #5
wat2000 said:
so am I subtracting every time the ball bounces?

Not subtracting, but adding.
If you drive your car for instance forward and backward and forward and backward, the distance meter will keep going up.
 
  • #6
I think they are looking for total distance, not net displacement.
 
  • #7
If you drive your car for instance forward and backward and forward and backward, the distance meter will keep going up.

Thanks for making me have to drive around the block backwards. I did not believe you, having watched Ferris Bueller about infinity times, but now I do--you are correct (for an '08 Jetta, anyway).
 
  • #8
lewando said:
Thanks for making me have to drive around the block backwards. I did not believe you, having watched Ferris Bueller about infinity times, but now I do--you are correct (for an '08 Jetta, anyway).

:rofl:
I watched Ferris Bueller too!

[edit]However, when the car went down, instead of forward or backward, I think the distance meter may have stopped going up :)[/edit]
 
  • #9
For example, the first bounce will come up to a height of 1 ft. When the ball falls back down and hits the ground for the second time, it will have traveled (3+1 +1) ft = 5 ft. Then the ball will rise again up to a third of a foot, and fall again. So the total distance once the ball has hit the ground for the third time is (3 + 1 + 1 + 1/3 + 1/3)ft. Continue this process until the ball has hit the ground for the fourth time.
 
  • #10
So would it be 3+1+1+1+1/3+1/3+1/3?
 
  • #11
wat2000 said:
So would it be 3+1+1+1+1/3+1/3+1/3?

not quite. Why is there three 1's in your sum? And why is the last term 1/3?
What is a third of 1/3? Try to visualize this. Draw a diagram.
 
  • #12
Down 3 feet.
Up 1/3 of that, 1 foot.
Down 1 foot. (total for this "up-down", 2 feet)
Up 1/3 of that, 1/3 foot
Down 1/3 foot. (total for this "up-down", 2/3 feet)
Up 1/3 or that, 1/9 foot
Down 1/9 foot. (total for this "up-down", 2/9 feet)

etc. (Why did you stop after only a few bounces- a geometric series can be summed for an infinite number of terms.)

(Ah, because the problem said "at the instant it hits the ground the fourth time", of course.!)
 
Last edited by a moderator:
  • #13
Here is the sequence if you are interested :P

[tex] 3 + \sum_{n=0}^{2}{\frac{2}{3^n} [/tex]

Or perhaps a more insightful version of this is:

[tex] 3 + \sum_{n=0}^{2}\left({\frac{1}{3^n} + \frac{1}{3^n}}\right) [/tex]

Or for an arbitrary number of bounces, the distance after the ball has hit the ground for the [tex] B^{th} [/tex] time is

[tex] 3 + \sum_{n=0}^{B-2}\left({\frac{1}{3^n} + \frac{1}{3^n}}\right) [/tex]
 
  • #14
Does this work? d1=3
second time d2=3*K=1
Third time d3=1*k=1/3
Forth time d4=K(1/3)=1/9

Now adding up d1+d2+d3+d4, we get the total distance
d=40/9
 
  • #15
ok the answer is 53/9!
 
  • #16
js14 said:
Does this work? d1=3
second time d2=3*K=1
Third time d3=1*k=1/3
Forth time d4=K(1/3)=1/9

Now adding up d1+d2+d3+d4, we get the total distance
d=40/9

Nope. Did you not read HallsofIvy's post?

HallsofIvy said:
Down 3 feet.
Up 1/3 of that, 1 foot.
Down 1 foot. (total for this "up-down", 2 feet)
Up 1/3 of that, 1/3 foot
Down 1/3 foot. (total for this "up-down", 2/3 feet)
Up 1/3 or that, 1/9 foot
Down 1/9 foot. (total for this "up-down", 2/9 feet)

EDIT: Re post #15: Now you got it!
 
  • #17
I got that part of the answer but now it asks this: Find a formula for the total distance the ball has traveled at the instant it hits the ground the nth time. I solved the problem like in the above problems. Is 9-(1/3)^n-3 the right formula?
 
  • #18
You mean this?
[tex]9 - \left( \frac{1}{3} \right)^n - 3[/tex]

If so, then that is wrong.
 
  • #19
no 9-(1/3) raised to the n-3.
 
  • #20
wat2000 said:
no 9-(1/3) raised to the n-3.

this is not right.. if summed from n=1 it gives

[tex] 9 - 6 - 3 - ... [/tex]

it will end up being a negative distance...

The correct sequence was shown in an earlier post.
 
  • #21
Yea I saw the formula with the sigma notation but my online homework dosent have sigma signs so their must be another way to write the sequence using only fractions...?
 
  • #22
let G.S be a Geometric Sequence where a=3 r=1/3
S4=(3*((1/3)4-1))/((1/3)-1)=(40/9)
 
  • #23
kiko_milzo said:
let G.S be a Geometric Sequence where a=3 r=1/3
S4=(3*((1/3)4-1))/((1/3)-1)=(40/9)

But 40/9 is not the correct answer
 
  • #24
I guess no one knows the correct answer...?
 
  • #25
js14 said:
I guess no one knows the correct answer...?

[tex] 3+1+1+1/3+1/3+1/9+1/9 [/tex]

The correct answer has been shown several times in this thread.
 
  • #26
and the sequence goes, (after starting at 3)

[tex] \frac{2}{3^n} [/tex]

starting from n = 0

But the formula for the total distance must be a summation! A sequence is just a lot of numbers...

Once the ball has hit the ground for the nth time, it will have traveled a distance D

[tex] D = 3 + \sum_{i=0}^{n-2}{\frac{2}{3^i} [/tex]
 
  • #27
I understand that answer its definitely correct, however my online homework has no sigma signs(meaning i can enter an answer like the one you show) so i was just suggesting that there must be a different way to write the equation. For example there is a similar question in my precalc book with the being dropped from 9 ft instead of 3ft and you were supposed to find the 5th bounce not the 4th. anyway the answer to the first part of the question was 17 (8/9) written as a mixed number. The equation to the problem(which is what I am trying to find out for this problem) was 18-(1/3)^n-3. no sigma notation. So what I am trying to explain is that i have to figure out how to put the equation to my problem into something similar to that. No sigma.
 
  • #28
js14 said:
The equation to the problem(which is what I am trying to find out for this problem) was 18-(1/3)^n-3.

Please, please, please, be precise with your notation. It's quite annoying when parentheses aren't used, because again, what you wrote looks like
[tex]18 - \left( \frac{1}{3} \right)^n - 3[/tex]
to me. And based on your previous post, I suspect you meant
[tex]18 - \left( \frac{1}{3} \right)^{n - 3}[/tex]
. You should really learn LaTex. If you cannot, at least for goodness sake use parentheses, like this:
18-(1/3)^(n-3)

Sorry if I sound cross. This is one of my pet peeves. Anyway, back to your question...

If you want to write out the sum of a finite number of terms in a geometric sequence, well, kiko_milzo, hinted at this, even though the answer wasn't correct to your original problem:
kiko_milzo said:
let G.S be a Geometric Sequence where a=3 r=1/3
S4=(3*((1/3)4-1))/((1/3)-1)=(40/9)

You'll need the sum of the first n terms of a geometric series formula:
[tex]a + ar + ar^2 + ... + ar^{n-1} = \frac{a(1 - r^n)}{1 - r}[/tex]
 
  • #29
heres an idea.
Evaluate the sum that I provided above for different values of n.
You should find a series:
[tex] 3, 5, 17/3, 53/9, 161/27, 485/81, [/tex]
etc..
Do you see a pattern?
Perhaps its better to write these as mixed numbers. Then the pattern should be quite clear.
 

1. What is a geometric sequence?

A geometric sequence is a sequence of numbers where each term is found by multiplying the previous term by a constant value. This constant value is called the common ratio.

2. How do I find the 4th bounce in a geometric sequence?

To find the 4th bounce in a geometric sequence, you can use the formula an = a1 * rn-1, where an is the term you want to find, a1 is the first term in the sequence, and r is the common ratio. Simply plug in the values and solve for a4.

3. What are the steps to finding the 4th bounce in a geometric sequence?

The steps to finding the 4th bounce in a geometric sequence are:
1. Identify the first term, a1.
2. Determine the common ratio, r.
3. Use the formula an = a1 * rn-1 to solve for a4.
4. Substitute the values of a1 and r into the formula.
5. Simplify the equation to find the 4th bounce, a4.

4. Can I use a calculator to find the 4th bounce in a geometric sequence?

Yes, you can use a calculator to find the 4th bounce in a geometric sequence. Most scientific calculators have a function for finding the nth term in a geometric sequence. Simply input the values for a1 and r, and then enter 4 for n to find the 4th bounce.

5. How can I use the 4th bounce in a geometric sequence to solve a real-life problem?

The 4th bounce in a geometric sequence can be used to solve real-life problems involving growth or decay. For example, if you are measuring the population of a certain species that doubles every year, you can use the 4th bounce to predict the population after 4 years. You can also use it to calculate future values of investments that have a fixed interest rate.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
4
Views
3K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Precalculus Mathematics Homework Help
Replies
2
Views
3K
  • Advanced Physics Homework Help
Replies
8
Views
2K
  • Precalculus Mathematics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
34
Views
596
  • Precalculus Mathematics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
3K
  • General Math
Replies
2
Views
649
Back
Top