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Calculate the athletes's velocity

  1. Oct 20, 2007 #1
    A 55kg pole-vaulter falls from rest from a height of 5.0m onto a foam-rubber pad. The pole-vaulter comes to rest 0.3s after landing on the pad.

    a) Calculate the athletes's velocity just before reaching the pad.
    b) Calculate the constant force exerted on the pole-vaulter due to the collition.

    Thanks!!!
     
  2. jcsd
  3. Oct 20, 2007 #2

    Astronuc

    User Avatar

    Staff: Mentor

    We respectfully request that one show some effort and work with respect to the solution of the HW problem.

    Write the equation that shows the relationship between the change in gravitational potential energy and kinetic energy. Think conservation of energy.

    means constant acceleration.

    The pole vaulter decelerates constantly for 0.3s. Compare this with how long it took to fall 5.0 m.
     
  4. Oct 20, 2007 #3
    O.K. So this is what I have done so far:

    First we calculate the final velocity of the vaulter using kinematics:

    [tex]
    \[
    \begin{gathered}
    \Delta x = v_i \Delta t - \frac{1}
    {2}g\Delta t^2 {\text{ when }}v_i = 0{\text{ we've got }} \hfill \\
    \Delta x = - \frac{1}
    {2}g\Delta t^2 {\text{ and}} \hfill \\
    \Delta t = \sqrt {\frac{{2\left( {{\text{5}}{\text{.0}}} \right)}}
    {g}} = 1.0{\text{s}} \hfill \\
    {\text{Now we can find the final velocity using }}v_f = v_i - g\Delta t: \hfill \\
    {\text{a) }}v_f = - 9.81\left( {1.0} \right) = - 9.81{\text{m/s}}{\text{.}} \hfill \\
    {\text{b) }}F\Delta t = m\left( {v_f - v_i } \right) \Rightarrow F = \frac{{55\left( {0 + 9.81} \right)}}
    {{0.3}} = 1798{\text{N}}{\text{.}} \hfill \\
    \hfill \\
    {\text{But according to the back of the book both of them a wrong}}{\text{.}} \hfill \\
    \end{gathered}
    \]
    [/tex]

    Thanks!!!

    PS: Sorry about that.
     
  5. Oct 20, 2007 #4
    [tex]
    \[
    {\text{Got it }}v_f = \sqrt {2gh}
    \]
    [/tex]
     
  6. Oct 20, 2007 #5
    depending on sig figs your velocity might be wrong.
    also you could have gotten it faster using:
    [tex]v_f^2=v_i^2+2a\Delta x[/tex]

    you also switched the values in your formula for average force:
    [tex]F_{net}=\frac{m}{\Delta t}(v_f-v_i)[/tex]

    also your values for the final/initial velocity are wrong, you want final/initial velocity for the collision. -9.81 m/s will be your Initial velocity (because that is the speed at which the pole vaulter approaches the floor). You need final velocity, you know it takes .3s you know the acceleration it's experiencing so you can find the final velocity and plug that in.

    Edit: hm..actually I don't think you need to do this last part, you only need to do it when the object bounces back up. your signs are just wrong in your final answer.
     
    Last edited: Oct 20, 2007
  7. Oct 20, 2007 #6
    sig figs issues :cry:

    Thank you!!!
     
  8. Oct 20, 2007 #7

    Astronuc

    User Avatar

    Staff: Mentor

    Yes. Initially, one can use mgh = 1/2 mv2, since the object is in freefall from rest, so all the gravitational potential energy is converted to kinetic energy, and given the same mass, then gh = v2/2.

    Slowing down is faster. Starting at v, the mass slows over [itex]\Delta{t}[/itex] for an average acceleration (or deceleration) of a = [itex]\Delta{v}[/itex]/[itex]\Delta{t}[/itex] = v/[itex]\Delta{t}[/itex].

    Please refer to

    http://hyperphysics.phy-astr.gsu.edu/hbase/mot.html

    and also

    http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html
     
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