High Dive Unit Problem:The Farrell Ferris Wheel

By Morgan Ashley ChangMU Block

Problem Statement

The unit problem for the High dive unit brings us to the meeting where the CAT seniors are planning out the CAT-nival, an event they are hoping will bring in large amounts of support that will fund the CAT graduation. Since no carnival would be complete without a major stunt, Dakarai, Leah, and John C. devise the perfect crowd pleaser.

The finale of the night will showcase Andre, Sheryl and Kaid on the Farrell-is wheel. Their planned stunt will be to drop Andre off of the moving ferris wheel into a pool of water below. At the end of the problem, we will have calculated out exactly how long Andre should be on the ferris wheel before he dives, so he will stay safe and hit the water perfectly instead of the ground below.

We began by looking at the speculations for the Farrell-is wheel and the dive to better understand all of the numbers that would be going into our calculations. We were provided with the information that:

- The radius of the ferris wheel is 50 feet- The center of the ferris wheel is 65 feet off the ground- The ferris wheel makes a complete turn in 40 seconds, which means it has a constant

angular speed of 9 degrees per second- The ferris wheel turns counterclockwise- The pool is attached to a moving cart- The moving cart starts 240 feet to the left of the ferris wheel- The moving cart has a constant speed of 15 feet per second- The surface of the water in the pool is 8 feet above the ground - When the cart and ferris wheel begin moving, the platform is at the 3 o’clock position

Though this initial information was very useful, there was still more information needed before we could begin to solve for the final time, such as:

- How long will it take for Andre to fall from the platform into the pool?- How high in the air is Andre when he begins his fall?- Does Andre’s height affect any of the equations?- What about Sheryl or Kaid’s heights?- What about Andre’s weight?

The Plan

- I will use the final equation in my tool kit to solve this problem, as it is a combination of all of the equations and expressions we have discovered throughout the unit.

- I will use the guess and check method to find the ideal time that Andre should be dropped at.

- (Added after beginning work) I will graph the two sides of the equation, where the intersection will provide me with the exact time.

- I will plug the final number back into the final equation to double check that it is the exact moment, in fear that any number too broad could be detrimental to Andre’s health.

Work

The first thing I did before even attempting to solve the unit problem was to get my remaining questions answered.

- How long will it take for Andre to fall from the platform into the pool? This was something I would find out by the end of my work, after having found all the correct equations needed to solve the problem.

- How high in the air is Andre when he begins his fall? I would find an equation for this as well.

- Does Andre’s height affect any of the equations? No, this did not need to be taken into consideration.

- What about Sheryl or Kaid’s heights? This was also of no importance to our problem.- What about Andre’s weight/mass? I soon learned in physics class that mass does not

affect one’s velocity when falling, so this point was now moot.

After completing all of the assignments, I now know the final equation needed for solving the unit problem is

However, we were not given this equation from the get go, and needed to formulate it as a culmination of all equations or expressions found throughout the unit.

The first equation we found was how to find the height of Andre at any given time (w).This equation was

.

The h(t) stood for “height at any given time”, which would be represented here as “t”, but later as “w”. The 57 was representative of the distance from the center of the ferris wheel to the top of the water in the pool. We found the number 57 because the center of the ferris wheel was 65 ft off the ground, but we had to account for the fact that we wanted to hit the top of the pool, and not the ground. To do this, we subtracted 8ft (the distance from the top of the water in the pool to the ground) from 65ft (the distance from the center of the ferris wheel to the ground and got 57 ft. The 50 represented the radius of the ferris wheel. And the 9 represented the angular speed of the ferris wheel, telling us that the ferris wheel moves 9 degrees per “t”, or second passed. when we combine the radius of the ferris wheel (50), and the angular speed of the ferris wheel (9) with the sine function, we are able to see the y-value or y-coordinate because of the unit circle.

The unit circle is a tool or concept used in trigonometry that helps visually show how sine and cosine functions relate with respect to right triangles. The unit circle has a radius of only 1 unit, which is why it is referred to as the unit circle. When we use the unit circle, we use the center origin of (0,0) and the appropriate degree we are looking for to find either the x or y coordinate. The sine is represented by the y-value on the unit circle, since the sine function is defined as opposite over hypotenuse (opp/hyp). However, we already know the hypotenuse will always be 1, since the radius of a unit circle is always 1 unit. The same goes for the cosine function, though it is defined as adjacent over hypotenuse (adj/hyp), which means that the value found with cosine gives us the x-value

instead. An important thing to remember when using the unit circle is that sine’s y-coordinate increases when the degrees increase, while the cosine’s x-coordinate decreases when the degrees increase.

Reviewing back to our first equation that gives us Andre’s height at any time, we know we are looking for a spot along the y-axis, his y-value, which confirms that we need to be using the sine function, since that is what will provide us with the y-value.

The second equation that came to us that was of importance was the equation of how long it would take Andre to fall.

or

This equation provided us with “t”, the time it would take for him to fall a certain height, and factored in the number 16 as the pull of gravity, and the top equation being the height at which the fall begins, which was our first equation.

The next equation we found was how to find Andre’s horizontal position on a graph.

Similar to how we use the equation for his height, the main difference in this equation is the use of the cosine function rather than the sine function. We use the cosine function because it provides us with the x-value, therefore x-coordinate for Andre at the time of his fall.

After this equation, it is important to be able to find the cart’s horizontal position in terms of falling time and the time spent on the wheel. The equation is:

) + w]

The -240 represents the starting position of the cart, or its x-coordinate at the time where t=0. We get the 15 by knowing that the cart travels at a constant speed of 15ft/s. So 15t would give us the total distance traveled after “t” amount of seconds. From previous equations, we know that “t” is made up of f+w, also known as the time spent falling, and the time spent on the wheel. Wheel time is simply represented by “w”, while the falling time is represented by the equation we found earlier in the unit.

Answer

ATTN: CAT-nival Committee

Dear CAT-nival Committee,

After many careful calculations, I have found the exact time that Andre must be dropped from the platform on the Farrell-is wheel to ensure he meets the pool safely. The time at which Andre needs to be dropped, down to the millionth of a second is 12.282855 seconds.

To find this answer, I used the final equation we discovered:

To review, the different numbers & variables in the equation account for:a) the cart’s starting positionb) the speed the cart travels atc) The height of Andre and the platform at the time which he begins to falld) gravity’s effect on Andre’s free falle) the amount of time Andre has spent on the wheel before he begins his dropf) the radius of the ferris wheelg) the angular speed of the ferris wheel

The time of 12.282855 seconds is the only time which is 100 percent exact and will ensure Andre’s safety as long as no other factors such as extreme winds or a malfunction with either the ferris wheel or the moving cart.

I can assure you of this because after I was able to successfully graph the two sides of the equation, and located their intersection at 12.282855 seconds, I plugged the number back into the equation to see if it indeed fit.

I ended up with:

-17.54770575 -17.54771

Though there is a 0.00000425 second difference between the two time, there is nothing to worry as the average human reaction time is anywhere from 0.2 to 0.25 seconds, so Sheryl & Kaid would not be able to react to a 0.00000425 second difference in time. This said, I still believe that 12.282855 seconds is the best time at which Andre should be released.

Reflection

This unit’s problem was difficult to complete, though the issue was not with solving the problem itself. The part in which I struggled with the most was compiling all of my work and thoughts together to finish the final typed report. I have always believed that the most challenging aspect about math is the ability to explain to others what you have figured out in your head, and this project was no different.

Out of my four years of math at CAT, I can whole heartedly say that the use of a tool kit has been most beneficial to me during this unit versus all previous ones. The requirement that we all needed to be signed off on a completed portfolio of past classwork, homework and warm ups forced me to reevaluate my organization habits when it came to keeping my math work in order. After spending a whole class either redoing work, or looking through every crevice and folder in my backpack and binder for a warmup I knew I had done, I decided to redo my tool kit entirely. I did this with the knowledge that even if I were to lose some papers again, I would have all the necessary information on one sheet of paper that would be a slight bit harder to lose due to its green color.

Sadly, this realization that an organized tool kit could make life easier did not come to me until right before we began our unit problem work, and long after we had taken our quizzes. I had stated in a previous unit problem that studying combinations and permutations was the hardest unit I had come across. I’m now retracting that statement and instead saying that this unit has been the most difficult for me to get through.

In retrospect, I am not sure if dealing with periodic functions was necessarily easier or more difficult than dealing with combinations and permutations, but I think the real issue was more of a mental distinction. When we embarked on combinations and permutations, we were given a warning that it would be the turning point of our math class this year, a warning that the work we’d be doing was about to be on a different caliber than that of our IMP 3 days. This in mind, I felt it was okay to not understand the concepts as easily the first time around.

However, during this unit, something in my mind seemed to have me thinking that it would be easier than the previous one. Perhaps I had convinced myself that if I had gotten over the last unit, I could surely do anything within the math world now. Whatever the reason was, I struggled with this unit just as much, if not more than during “What are the Odds?” I was hit most effectively in my quiz grades, which came back showing that I had not retained the knowledge nearly as well as I had imagined, and that not even my tool kit could save me from this lack of understanding (though a poorly put together tool kit can be attributed with some of the blame as well).

But I forced myself to not dwell too much upon the low quiz grades, and told myself to focus on the unit problem instead, which I was sure could help redeem my grade if the work was done well. Still, I found myself getting a slow start on the work, which I was hoping to avoid for once in my high school career.

After we began to work on our problem, we were assigned to complete the problem statement over the weekend, which I did like a good little student. I however, did not do anything past this, which was a bad choice. I was absent to the following class, which put me way behind the other students, who had used that class to start on their

work sections, and had went on to solve the problem. Although there was nothing I could’ve done about my absence, I regretted not having worked on ahead over the weekend.

Luckily, the day I came back, there was still a Unit Problem Party after school, which I am very glad to have attended. I normally do not make it to the POW parties or unit problem parties because they’re usually held before school. I was lucky enough that this time, the “party time” had worked out in my favor, and I was able to get a lot of work completed that I could have never gotten done at home or on my own.

When I sat down to begin my work, I tackled it in accordance to my plan, using guess and check. The first number I attempted was 20 seconds, as it was half the time it takes for a complete revolution, and would place Andre at the 9 o’clock position, which seemed like a very feasible and simple place for a jump. I should’ve known it wasn’t going to be that easy.

After testing 10 seconds, then 15, followed by 12.5, 11, 12, and 12.3, I began to get stumped. 12.3 seconds seemed to be the closest I could get while staying in the tenth decimal range. I knew I would need to get more decimals to more accurate, but after testing all of these number, I was not sure which way to go in terms of guessing. Should I attempt 12.31 seconds next, or 12.29 seconds?

Fortunately, the most important benefit from being at the party was receiving help from Ariel, who helped steer me towards a more precise answer, which I couldn’t have been more grateful for. Ariel came by to chat me up, and saw that I was struggling.

“Have you found the answer yet?” she asked, to which I responded with a frustrated “I’m this close.”

She looked at my work before exclaiming, “Wait, you’re not graphing it?” “No…” I began. “Ohmygosh! Graphing is so much easier!”I thought about it, and proceeded to thank her immensely as I realized just how

right she was. How had I not seen that graphing the two sides of the equation, and finding their intersection would be the surefire way to go?

“Just be careful about the parentheses,” she advised, “the calculator is kinda stupid, and doesn’t know how to follow out PEMDAS.”

Sadly, she was completely right, and it took me the rest of the night to get the parentheses perfectly placed and send the “syntax error” screen away for good. After I was able to successfully graph the two sides of the equation, and located their intersection at 12.282855 seconds, I plugged the number back into the equation to see if it indeed fit.

I ended up with: -17.54770575 -17.54771

I subtracted the smaller time from the larger time and found that there was a difference of 0.00000425 seconds between the two times. Though this slight difference in numbers agitated me at first, and had me desiring to go back and find an even more precise number, I took another route instead. I used the internet to research what the common human reaction time was, and according to an online version of a physics textbook, the average human reaction time is anywhere from 0.2 to 0.25 seconds. Relieved that the average human reaction time was slow enough so that a difference of

0.00000425 seconds would not matter, I settled with my answer and went on to work on the rest of the report.

Looking at the final equation that was used to solve the problem, there are four of the seven variables that, if altered, could change the outcome of the entire problem. The first of these variables is the cart’s starting position. If the cart were to begin at a lower number, such as -300, it would be further to the left of the ferris wheel, which would give it a greater distance to travel. Because of this, Andre would need to stay on the ferris wheel longer before the point of intersection between both graphs.

Another variable whose change could affect the final outcome is the speed at which the cart travels. If the cart moves at a faster speed, it would arrive at the original point of intersection earlier than the ferris wheel, and therefore surpass that original point. We can see on a graphing calculator if we were to change the speed of the cart from 15ft/s to 20ft/s, the point of intersection would come sooner, and would be closer to the 10 second region, rather than the 12 second region.

The next two variables whose numbers affect the final outcome deal with the ferris wheel instead of the cart. The first of these two is the radius of the ferris wheel. The radius plays an important role as it affects Andre’s position at any given time. Using the graphing calculator again, if we change the ferris wheel’s radius from 50 ft to 75 ft, the time needed to stay on the ferris wheel is reduced from around the 12 second region to the 11 second region. This happens because the furthest edge of the ferris wheel now extends outwards more, so the 3 o’clock and 9 o’clock x-coordinates are now at 75 and -75 rather than 50 and -50. This means that there is less of a distance between the cart’s starting point and the point where it crosses into the ferris wheel’s area, so there is less time needed before the two graphs intersect.

The last variable that can affect the final outcome is the angular speed of the ferris wheel. If the ferris wheel instead moves at a slower pace, such as 5 degrees per second versus 9 degrees per second, it will require a longer amount of time before the platform even crosses into the left quadrants of the graph, where it can meet with the moving cart.

After thinking about how this problem can be visually represented, it occurred to me just how similar it was to the water balloon dropping experiment we had done in Mr. Richard’s class. Mr. Richards represented the moving cart, as we were given his starting position and his constant walking speed. Our water balloon served as Andre, who we aimed to meet with Mr. Richards at an exact time. The only difference was the fact that we were not on a moving ferris wheel, so our “platform” in the third floor was stationary, which made the problem a bit simpler. Knowing how to solve the unit problem could have helped us out in this experiment, hopefully allowing at least one group to successfully hit Mr. Richards with a water balloon.

Ultimately, I am glad to be done with this unit problem, as it forced me to understand trigonometry at a level I had never anticipated or sought out. I am fortunate enough that the problem was posed in a playful manner, or else I’m not sure I would’ve had the motivation to go through with solving it, as numbers without meaning are harder to deal with. However, I was able to deal with these numbers, and the sense of accomplishment couldn’t be more relieving.