# Rebound Superball Problem

• Dec 31st 2009, 11:53 AM
bkap
Rebound Superball Problem
The rebound ratio of a speckled green superball is 75%. It is dropped from a height of 16 feet. Consider the instant when the ball strikes the ground for the fiftieth time.
a. How far downward has the ball traveled at this instant?
b. How far ( upward and downward has the ball traveled at this instant?
c. How far would the ball travel if you just let it bounce...?

I have a feeling for the first two it would be something along the lines of "16(0.75)^0+16(0.75)^1...+16(0.75)^50", and for the second 2(" "), but there must be something that would allow me to do this calculation quicker than just adding 50 numbers together. Would it have something to do with ! ?

Thanks.
• Dec 31st 2009, 12:22 PM
sym0110
Quote:

Originally Posted by bkap
The rebound ratio of a speckled green superball is 75%. It is dropped from a height of 16 feet. Consider the instant when the ball strikes the ground for the fiftieth time.
a. How far downward has the ball traveled at this instant?
b. How far ( upward and downward has the ball traveled at this instant?
c. How far would the ball travel if you just let it bounce...?

I have a feeling for the first two it would be something along the lines of "16(0.75)^0+16(0.75)^1...+16(0.75)^50", and for the second 2(" "), but there must be something that would allow me to do this calculation quicker than just adding 50 numbers together. Would it have something to do with ! ?

Thanks.

Yes your approach to the first problem is correct, if you factorize things out a bit, it looks something like:
$\displaystyle 16*((3/4)^0+(3/4)^1+...+(3/4)^{49})$
it ends with the 49th power because 0-49 gives a total of 50 drops
now this is just a standard progression, whose sum is:
$\displaystyle a(r^{n+1}-1)/(r-1)$
where a is the first term, in this case 1, r is the successive ratio, in this case 3/4, and n is how far the last term goes, in this case 49.

The second part is trivial once you got the first part, however you have to keep in mind that the first upward bounce is the same distance as the second downward bounce.

Finally for the third part, you simply let n tend to infinity and calculate. If you haven't done limit, just make n=1000 on the calculator, this will give you a pretty good approximation
• Dec 31st 2009, 12:24 PM
e^(i*pi)
Quote:

Originally Posted by bkap
The rebound ratio of a speckled green superball is 75%. It is dropped from a height of 16 feet. Consider the instant when the ball strikes the ground for the fiftieth time.
a. How far downward has the ball traveled at this instant?
b. How far ( upward and downward has the ball traveled at this instant?
c. How far would the ball travel if you just let it bounce...?

I have a feeling for the first two it would be something along the lines of "16(0.75)^0+16(0.75)^1...+16(0.75)^50", and for the second 2(" "), but there must be something that would allow me to do this calculation quicker than just adding 50 numbers together. Would it have something to do with ! ?

Thanks.

Nah, Factorials only really occur in probability - you'll be better off with a sequence. Let s be the distance it bounces back up

$\displaystyle s_0: 16 = 16 \times 0.75^0$
$\displaystyle s_1: 16 \times 0.75^1 = 0.75s_0$
$\displaystyle s_2: (16 \times 0.75) \times 0.75 = 0.75s_1$
...
$\displaystyle s_n = 16 \times 0.75^n = 0.75s_{n-1}$

From this we can see it's a geometric sequence so use the sum of a geometric sequence to solve how far it's bounced up. To find down and up multiply by 2
• Dec 31st 2009, 01:19 PM
Soroban
Hello, bkap!

Everyone has given you excellent advice . . .

Quote:

The rebound ratio of a speckled green superball is 75%.
It is dropped from a height of 16 feet.
Consider the instant when the ball strikes the ground for the 50th time.

a. How far downward has the ball traveled at this instant?

Total downward distance is given by:

. . $\displaystyle D \;=\;16 + 16(.075) + 16(.075)^2 + 16(0.75)^3 + \hdots + 16(0.75)^{49}$

. . . . . $\displaystyle =\;16\underbrace{\bigg[1 + (0.75) + (0.75)^2 + (0.75)^3 + \hdots + (0.75)^{49}\bigg]}_{\text{geometric series}}$

. . The sum of the geometric series is: .$\displaystyle \frac{1-(0.75)^{50}}{1-0.75} \:=\:3.999997735$

Therefore: .$\displaystyle D \;=\;16(3.999997736) \;=\;63.99996376$ feet.

Quote:

b. How far ( upward and downward) has the ball traveled at this instant?
We must be very careful . . .

The balls falls 16 feet.

Then bounces up $\displaystyle 16(0,75)$ feet and falls $\displaystyle 16(0.75)$ feet.

Then bounces up $\displaystyle 16(0.75)^2$ feet and falls $\displaystyle 16(0.75)^2$ feet . . . and so on.

The total distance is:

. . $\displaystyle T \;=\;16 + 2\!\cdot\!16(0.75) + 2\!\cdot\!16(0.75)^2 + 2\!\cdot\!16(0.75)^3 + \hdots + 2\!\cdot\!16(0.75)^{49}$

. . . .$\displaystyle =\; 16 + 32(0.75)\bigg[1 + (0.75) + (0.75)^2 + \hdots + (0.75)^{49}\bigg]$

. . . .$\displaystyle =\; 16 + 24(3.999997735) \;=\;111.9999275$ feet.

Quote:

c. How far would the ball travel if you just let it bounce?
If we let it bounce (forever), the total distance is:

. . $\displaystyle T \;=\;16 + 24\underbrace{\bigg[1 + (0.75) + (0.75)^2 + (0.75)^3 + \hdots\bigg]}_{\text{infinite geomtric series}}$

. . The sum of the infinite series is: .$\displaystyle \frac{1}{1-0.75} \:=\:4$

Therefore: .$\displaystyle T \;=\;16 + 24(4) \;=\;112$ feet.