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Trying to find an angle

  1. Aug 18, 2003 #1
    Here it goes:

    After bungee jumping off a bridge, a person begins to swing as a pendulum, the rope being 77 m long. The person's speed is 14 m/s as it passes through its lowest point. (a) find the height to which it rises above this position before stopping. (b) what angle does the pendulum then make with the vertical?


    L = 77 m
    Vmax = 14 m/s

    T (period of pendulum) = 2 * pi* squareRoot(L / g)

    T = 2 * pi * squareRoot(77 m / 9.8 m/s^2)

    T = 17.6 s

    Avg Velocity = (Vf - Vi) / 2
    = (14 m/s - 0) /2
    = 7 m/s
    dt = avg Velocity * t
    = 7 m/s (17.65 s)
    = 1232 m
    d = dt / 4
    d = 30.8 m

    My problem is finding the angle. because the distance is not straight line so there cannot be a right triangle. I have absolutely no idea what to do here. Am I right on the first part. I divided dt by 4 because there is 4 parts to the period of a pendulum. At least I think thats right.
  2. jcsd
  3. Aug 18, 2003 #2


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    Science Advisor

    Why are you calculating an "average" velocity? The problem is assuming that the person has already jumped, has reached the bottom- at the end of the rope- and is now swinging as a pendulum.

    The way I would do this is use "conservation of energy". You know the person's speed at the bottom so you can calculate the kinetic energy. You know that, at the top of the swing, the person's speed is 0 so all that kinetic energy is converted to potential energy. That allows you to find how much above the bottom he is (call that x). You now have a right triangle with hypotenuse of length 77m, near leg of 77- x so that the angle is given by arccos((77-x)/77).
  4. Aug 24, 2003 #3
    I think Ivy is right - you can say that when a pendulum is at its greatest height, its kinetic energy has all been converted into potential energy. KE = mv2/2 ; PE = mgh. if you equate these two and solve for h, you get the height where the pendulum stops, a smaller height than 30 m (this distance is naturally relative to the lowest position of the pendulum).
  5. Oct 21, 2003 #4
    I have a problem similar to this. In fact, it's the same only with different values:

    A ball at the end of a 180-cm long string swings as a pendulum as shown. The balls speed is 400 cm/s as it passes through its lowest position. (a) To what height h above this position will it rise before stopping? (b) What angle does the pendulum then make to the vertical?

    I think my big problem is that I don't get the concepts behind it yet. What principles are at work here?

    What I have:

    The period (time it takes for the pendulum to move from lowest position to highest) is figured by:

    2 * pi sq(L/g)

    Since (a) asks for h, there is a conservation of momentum aspect to the question, but will mv2/2 = mgh work in this situation?

    m's cancel
    v=4 m/s
    g=9.8 m/s^2

    (4^2)/2 = 9.8h
    h=0.816 m

    Is that right? It seems... too easy. And weird.

    And as for (b), this needs a right triangle, right?

    The hypotenuse would be L and the side on the other side of theta would be L-t, and the other side is irrelevant since we can get theta w/ the cosine.

    So then it would be cos(theta)=(L-h)/L

    So... cos(theta) = (1.8 - 0.816)/1.8
    theta=56.9 deg

    Um... is that right?
    Last edited by a moderator: Oct 21, 2003
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