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Math Help - free falling.

  1. #1
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    free falling.

    A student (m = 63 kg) falls freely from rest and strikes the ground.

    During the collision with the ground, he comes to rest in a time of 0.0200 s.

    The average force exerted on him by the ground is +18500. N, where the upward direction is taken to be the positive direction.

    From what height did the student fall? Assume that the only force acting on him during the collision is that due to the ground.
    in meters


    is this a problem solved using a kinematic equation?
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by rcmango View Post
    A student (m = 63 kg) falls freely from rest and strikes the ground.

    During the collision with the ground, he comes to rest in a time of 0.0200 s.

    The average force exerted on him by the ground is +18500. N, where the upward direction is taken to be the positive direction.

    From what height did the student fall? Assume that the only force acting on him during the collision is that due to the ground.
    in meters


    is this a problem solved using a kinematic equation?
    No, but that is a part of it.

    In order to use a kinematic equation, or an energy equation for that matter, we need to know how fast the student was falling just before impact. We can get that number by using the Impulse-Momentum Theorem. As the problem stated, I'm going to use upward as a +y axis.

    J = \Delta p

    \bar{F} \Delta t = p - p_0

    \bar{F} \Delta r = m(v - v_0)

    The final speed of the student is 0 m/s, so
    \bar{F} \Delta t = -mv_0

    So
    v_0 = - \frac{\bar{F} \Delta t}{m}

    So: \bar{F} = 18500~N, \Delta t = 0.0200~s, and m = 63~kg.

    Note that your v_0 will be negative. This is because + is upward and the velocity just before impact is downward. (This is a good check to see if our equation is right.)

    Once you get v_0 you can either do a work-energy or kinematics problem to determine the height.

    -Dan
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  3. #3
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    So, even though the problem says it starts from rest, I must assume the v0 is not 0, because it must have an initial JUST before it hits the ground which would be the initial velocity?

    ...okay so, v0 is -5.87



    so i used a kinematics equation of: d = ( (vi + vf) / 2) * t

    where d is:

    d =vi * t + 1/2 * a * t^2

    d = -5.87 + 1/2* (9.8)* .0200^2

    ...and that did not work.

    so i found t by using vf + vi/ a .... 0 + 5.87 / 9.8 = t

    t = .598979592

    okay so i tried to



    .....................nevermind its 1.76 meters =D

    i just used a different kinematics equation.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by rcmango View Post
    So, even though the problem says it starts from rest, I must assume the v0 is not 0, because it must have an initial JUST before it hits the ground which would be the initial velocity?

    ...okay so, v0 is -5.87



    so i used a kinematics equation of: d = ( (vi + vf) / 2) * t

    where d is:

    d =vi * t + 1/2 * a * t^2

    d = -5.87 + 1/2* (9.8)* .0200^2

    ...and that did not work.

    so i found t by using vf + vi/ a .... 0 + 5.87 / 9.8 = t

    t = .598979592

    okay so i tried to



    .....................nevermind its 1.76 meters =D

    i just used a different kinematics equation.
    This post sounds like you are using any equation you can think of to find an answer, even if it doesn't apply, and if it doesn't work you are moving on to the next one.

    You need to be careful about the definitions. Perhaps I erred when I used v_0 to represent the speed of the student just before collision. This v_0 is not the same as the initial velocity with which the student started to fall.

    We already have +y upward. I'm going to put an origin on the ground. So let's fill out a table.
    t_0 = 0~s
    y_0 = ? and y = 0~ m
    v_0 = 0~m/s and v = -5.87302m.s
    (This final velocity is the v_0 from the impulse-momentum part of the problem.)
    a = -g

    We are looking for y_0. Given the known and unknown information you can immediately see that
    v^2 = v0^2 + 2a(y - y_0)
    will do the trick.

    Plan out how you are going to do the problem, then organize it.

    -Dan
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