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Homework Help: Applying trigonometric functions to some real life situations

  1. Mar 6, 2006 #1
    I'm having some trouble with applying trigonometric functions to some real life situations, particularly this one problem in my homework.

    Andrea, a local gymnast, is doing timed bounces on a trampoline. The trampoline mat is 1 meter above ground level. When she bounces up, her feet reach a height of 3 meters above the mat, and when she bounces down her feet depress the mat by 0.5 meters. Once Andrea is in rhythm, her coach uses a stopwatch to make the following readings:

    1. At the highest point the reading is 0.5 seconds.
    2. At the lowest point the reading is 1.5 seconds.

    Now I'm not asking anyone to do my homework as it would defeat the whole purpose of me learning, but I would certainly love some advice as to how to formulate an equation and graph for this particular problem.

    One such problem I have with the problem is figuring out where the graph begins on the y-axis, or in other words, how high she is when the coach starts timing her.

    Any help would be greatly appreciated. Thanks.
    Last edited: Mar 6, 2006
  2. jcsd
  3. Mar 6, 2006 #2
    to begin this problem i would suggest that you draw four lines corresponding to each line the bottom is at ground level, @ 1m is the base of the trampoline, @2m is some height and @3m you have her highest point. Now draw a horizontal line for time, and mark off units of 0.5 corresponding to seconds. Finally, plot the given points, from the coaches readings, i.e. at t=0.5 she's at 3 m and at t=1.5 she's at 0.5 m. Now draw a 'cosine' like plot connecting the points and you've got a rough sketch of her motion as a function of time between the given intervals. hope this helps, sincerely, x
  4. Mar 6, 2006 #3
    That actually helps me a lot, but rather than drawing a cosine graph, I drew a sine graph. Thanks a bunch x.

    I've come up with an equation, but I'm not sure how right I am, if someone could verify, that would be great.


    The thing in the brackets is pi, sorry, but I don't know the keystroke for it.
  5. Mar 6, 2006 #4
    looks like you got it. good job, and i'm glad i could help. sincerely, x
    [tex] y(x) = 1.25 \sin(\pi x)+1.75 \Rightarrow y(0.5) =3, y(1.5) =0.5 [/tex]
    note also you have the period, amplitude and all right from the equation, this is important to note if you are doing a lot of graphing with trig func.
  6. Mar 6, 2006 #5
    Yeah, thanks a bunch x, I've figured out that question. However, I have a few other questions. This regards a ferris wheel problem I'm working with.

    Ferris wheel:
    Diameter: 76m
    Maximum Height: 80m
    Ferris wheel has 36 carts, which each cart able to hold approximately 60 people.
    It rotates once every 3 minutes.

    Work that I've figured out:
    Equation: [tex] h(t)=38\cos(2\pi/3)x+42 [/tex]
    (assuming that the ferris wheel begins rotating at maximum height)

    Now it asks me this:

    How many seconds after the wheel starts rotating does the cart first reach 10 meters from the ground?

    Now when I try doing this, my answer comes up negative, which isn't possible. Can anyone help me on this?
  7. Mar 6, 2006 #6
    sorry, i took so long to get back i didn't know you had another question. your equation for the equation of motion is correct, with your assumption the cart starts at y(t=0)=80 m. so you want to know when the cart will first reach 10 m from the ground.
    [tex] 10 = 38\, cos \left( \frac{2 \pi t }{3} \right) +42 \Rightarrow t=-\frac{3 cos^{-1} \left(\frac{16}{19} \right)-3 \pi}{2\pi} \sim 1.23 \, min [/tex]
    this is what i got, which makes sense, if the cart starts from the top then our time should be less than half the period since at that point our height above the ground is 4 m. Thus, starting from the top and rotating we first reach 10 m above ground approx 1.23 min later. hope this helps, x
    Last edited: Mar 6, 2006
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