Maya Wilson
Franklin and Marshall College
Lancaster, PA 17604
April 11, 2008
Iowa Jones
Fine Eye Times
475 Second Street
Big City, PU 11248-1632
Dear Iowa,
I am sorry that it has taken me so long to get back to you. I know this a very serious situation. However, I wanted to make sure that my answer did not put you or Will in any unnecessary danger. Your letter asked me if there was an interval of 9 seconds or more in which to enter a second password to save your beloved Will from the parabolic tank that Eve has been keeping him in. Now I write with great news! You can save Will! In fact, not only do you have a 9 second interval to enter the required password, but you have a whole 15 seconds.
I came to this conclusion through a series of calculations. The first thing I had to figure out was the time between bounces. After all, it was slightly possible that there would be a 9 second interval between when the first shoe bounced, went up to its maximum height, and then bounced again. To figure this out, I created a couple graphs using the information you gave me and analyzed them.
You told me that the
shoes were
initially dropped from a height of 100 feet, and that each bounce after
that
reached a height of 81% of the bounce before it. Knowing from the
various
physics books that I have read that gravity has an effect on any
falling
object, the acceleration of the shoes is 32ft/s2.
The piece of the second graph that I analyzed was the first half of the time it took for the shoe to go through the first full bounce. Therefore, T/2 represents the time it took for the shoe to reach its maximum height of that bounce. Since the height of the first full bounce was 81 feet (81% of 100 feet), I knew that the area under the velocity curve was 81. Also, since I knew the acceleration of the shoes was 32ft/s2, the slope of the velocity curve was 32. Since velocity can be found by multiplying acceleration by time, the velocity (height of the triangle) is 32T/2.
Knowing these pieces of the triangle, and that the area of a triangle is
A= ½ (base) (height),
I substituted the pieces of the triangle I was analyzing into the formula. When I did that, the formula became
81= ½ (T/2) (32T/2).
I then simplified this equation and found that T was 4.5 seconds. This means that the first full bounce took only 4.5 seconds. This was very unfortunate information. After all, the best chance at having a 9 second interval was during the bouncing of the first shoe for this first full bounce. Each bounce after that reached only 81% of the bounce before it, so the time it took for each full bounce gets smaller and smaller. Yet I was not ready to give up. There was still a chance to save Will.
It would seem that since the bouncing shoes behaved in this way of reduced height, the first shoe would never stop bouncing. However, I remembered a concept that would eliminate this worry. It is a mathematical formula that would allow the shoe to bounce an infinite amount of times in a finite amount of time. It involves finding a geometric series, which can be defined and explained in calculus textbooks. The idea is to add up all the times it would take for all the full bounces of the first shoe. If you can manipulate information to take the form of
1 + r + r2 + r3 + r4 + É
the sum of all these things is
Now all I had to do was manipulate what I knew to fit this form.
I did this by using the same graphs as before. However, instead of applying the graph to only the first full bounce, I applied it to all of the full bounces.
In the first graph, the height of each bounce is 81% of the bounce before it. Since the initial height is 100 feet, the general formula for the height of any bounce is
H = 100 (.81)n ,
where n is the number of the bounce. So the first bounce is 81 feet just as before, the second bounce is 65.61 feet, and so on. Now that I had the general formula for the height, I could plug it into the formula for the area of the triangle that I used before, since the area of the triangle was the height of the bounce. Replacing the 81 feet of height in the first calculation with 100 (.81)n, the formula for the nth bounce became
100 (.
I simplified the equation once more to find that
-2.5 + 5 +5(.9)1 + 5(.9)2 + 5(.9)3 + É .
When I take a 5 out of all but the first term, the sum becomes
-2.5 + 5(1 + (.9)1 + (.9)2 + (.9)3 + É ).
As you can see, the terms inside the parentheses fit the form of a geometric series. So the sum changes once more to
-2.5 + 5[1/ (1-.9)] = 47.5,
where r is .9.
This calculation shows that it would take 47.5 seconds from the time the first shoe initially dropped to the point where it stopped bouncing. You told me that the second shoe would fall 60 seconds after the first shoe fell. So there is a difference of 12.5 seconds between when the first shoe stops bouncing and the second shoe falls. However, there is also a time of 2.5 seconds from when the second shoe initially falls to when it hits the bottom of the tank for the first bounce. Therefore, there is a total time of 15 seconds between when the first shoe stops bouncing and the second shoe bounces for the first time.
I hope that my response was helpful. I came to my conclusions with the help of Professor Crannell, along with discussion with some of my classmates in calculus class, including Steve and Chris. If my final calculations are correct, and you are truly able to enter the second code within 9 (or 15) seconds, you will be able to save Will. Good luck and I hope all goes well!
Your friend,
Maya Wilson
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