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Force and Impulse when Landing

  1. Apr 4, 2013 #1
    1. The problem statement, all variables and given/known data
    When a person parachutes, the impact velocity is equal to that attained in free fall from a height of 4.5 m. After contacting the ground, the jumper’s momentum is quickly brought to zero by the earth. Assuming that the person has a mass of 80 kg, what is the average force on the jumper’s feet if he lands stiff-legged and the impulse lasts only 0.019 s?


    3. The attempt at a solution
    v22 = v12 + 2aΔd
    v22 = (2)(1.8)(4.5)
    v2 = 9.4 m/s [down]

    p = mv
    p = (90)(9.4)
    p = 752 N

    F = Δp/Δt
    F = 750/0.019
    F = 3.9 × 104 N [up]

    However, the correct answer is 4.03 x 104 N [up]. My teacher hinted towards the normal force to the person's weight is added to the answer. This confuses me as I thought that the impulse takes that into consideration already. Someone please help clarify this for me. Thanks in advance.
     
  2. jcsd
  3. Apr 5, 2013 #2

    NascentOxygen

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    Staff: Mentor

    Hi Killic... http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

    The answer you obtained (through omitting gravity) is the answer that applies if the problem was reworded for horizontal motion, viz., a speed skater comes off the rink at 9.4 m/sec and collides with a plasterboard-encased concrete wall, coming to a stop in 0.019 secs.

    But it takes additional force to halt a body moving vertically because gravity is trying to accelerate the body (pulling downwards) all the time that friction (pushing upwards) is trying to slow the body.
     
    Last edited by a moderator: May 6, 2017
  4. Apr 5, 2013 #3

    PeterO

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    Homework Helper

    Your teacher gave a good lead.
    You have calculated the net Force on the jumper.
    That is; the result after adding the upward force from the ground to the downward force due to gravity.
     
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