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One-Dimensional Kinematics: Height vs Time

  1. Sep 29, 2003 #1
    This problem has a picture. You can view it at http://www.geocities.com/rockdog_84/One.htm

    Ok. Maybe I'm missing something here, but for part one, I thought free-fall acceleration would be 9.8 m/s^2. Of course, computer didn't take it, so is there a way to find it from the graph?

    As for the second part of the question, I thought initial velocity is 0, but its not the right answer. :-(
    I guess another way to find initial velocity would be take the derivative of x(t) when t = 0, but I don't know the equation for this problem.

    Any insight is appreciated.
  2. jcsd
  3. Sep 29, 2003 #2

    Claude Bile

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    Science Advisor

    9.8 m/s^2 is the correct acceleration due to gravity on Earth, however the scenario in question takes place on the planet Zoot. If a question does not take place on Earth, then that is a big red flag that acceleration due to gravity is NOT 9.8 m/s^2.

    You can figure out g by using s = 1/2 at^2 + u*t (Take the second half of the graph where u is 0).

    Once you know g, you can use u^2 = v^2 - 2*a*s to find the final velocity. (Take the first half of the graph this time so v = 0)
    Note: -
    u = initial velocity
    v = final velocity
    t = time
    a = acceleration
    s = displacement

  4. Sep 30, 2003 #3
    big thanks!

    Man, I would never have thought of that! Thank you so much.
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