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Simple kinematic equations question?

  1. Nov 17, 2008 #1
    1. The problem statement, all variables and given/known data

    A 52.2 kg pole-vaulter falls from rest from a height of 3.6 m onto a foam-rubber pad. The pole-vaulter comes to rest 0.34 s after landing on the pad.

    (a) Calculate the athlete's velocity just before reaching the pad.

    (b) Calculate the constant force exerted on the pole-vaulter due to the collision.

    2. Relevant equations

    Kinematic equations: V = Vo + AT, X = ((V+Vo)/2)*T, V^2 = Vo^2 + 2AX, X = VoT +.5AT^2

    F = ma (though I haven't attempted part b yet)

    3. The attempt at a solution

    It seems like this problem should be easily and quickly solvable using any of the kinematic equations (4 of three values are known, right? Time, distance, acceleration, and original velocity), but this is what I tried and both answers were wrong:

    0 = Vo + 9.8 m/s^2 * .34 s (tried 3.33, didn't work)

    3.6 = Vo(.34) + .5 * 9.8 *.34^2
    3.6 = Vo * .19259 (tried 18.7, didn't work)

    I'm a little confused about whether the pole vaulter's mass is irrelevant information, or how to use it—I think I'm a little off with the kinematic equations. Thanks for your help and answers, I know I'm not very good at physics! :)
  2. jcsd
  3. Nov 17, 2008 #2


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    Homework Helper

    Hi sbacker,

    No, for part a you do not know the time. There are two parts to the motion: the fall from the highest point to the pad, and then the motion as the person pushes down into the pad (stopping).

    The 0.34seconds refers to how long it takes for the pad to stop the person, not how long it takes for the person to reach the pad.
  4. Nov 17, 2008 #3
    Looks like you've just misread the question. 0.34 seconds isn't how long it takes the pole vaulter to fall onto the pad, it's how long it takes the pad to "pop back" after the vaulter lands. So your first step should be to figure out how long it takes him to reach the pad. The 0.34 seconds don't matter until part b of the question.

    Edit: Oops. Thought I'd refreshed the thread before replying.
  5. Nov 18, 2008 #4
    Oh, I read the problem way too fast! How silly of me— thanks for clarifying, I got both parts of the problem right using the kinematic equations with only original velocity, acceleration, and distance. Thanks for pointing out my mistake!
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