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Vertical Velocity and Impulse

  1. Jun 12, 2009 #1
    1. The problem statement, all variables and given/known data

    This is an MCAT practice problem that is driving me crazy! The answer is supposed to be 2200N on each foot. Any help would be much appreciated! Thanks in advance!

    A guy, whose mass is 80 kg, jumps off a bench that is 1 meter off the ground. Immediately after landing on the ground with both feet, he jumps 1 meter up into the air. What is the average force on his left foot if the time of contact with the ground is 0.2 seconds?

    2. Relevant equations

    vf^2 = vi^2 + 2ad

    Fnet = m(vf-vi)/time

    3. The attempt at a solution

    Fnet = Fground - Fgravity
    Fgravity = mg

    Fground = Fnet + mg

    Need to find change in velocity:

    vf (at top of jump) = 0 m/s
    d = 1m (with downward chosen as positive direction)
    a = g = 9.8 m/s^2

    vf^2 = vi^2 + 2ad
    vi^2 = -2ad = -(2)(9.8 m/s^2)(1 m)
    vi = -4.43 m/s = 4.43 m/s upward

    Plugging into impulse equation:

    Fnet = m(vf-vi)/time
    Fnet = (80 kg) (4.43 m/s)/0.2 s = 1772 N

    Fground = Fnet + mg = 1772 N + (80 kg)(9.8 m/s^2) = 2556 N

    If the total Fground is 2556 N, the Fground on each foot is 2256 N/2 = 1278 N.

    As mentioned above, the answer is supposed to be 2200 N on each foot.
     
  2. jcsd
  3. Jun 12, 2009 #2

    LowlyPion

    User Avatar
    Homework Helper

    Welcome to PF.

    Isn't the change in velocity 2*Vf?

    Doesn't he have to rebound with the same velocity to achieve the same height again?
     
  4. Jun 12, 2009 #3
    Yes, of course, that makes perfect sense! Thanks so much!!
     
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