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Momentum of a Punt

  1. Feb 7, 2017 #1
    1. The problem statement, all variables and given/known data

    A punter drops a ball from rest vertically 1 meter down onto his foot. The ball leaves the foot with a speed of 18m/s at an angle of 55##^{\circ}## above the horizontal. What is the impulse delivered by the foot (magnitude and direction)?



    2. Relevant equations
    ##p_b+p_f=p^{\prime}_{b}+p^{\prime}_{f}##
    ##v^2=v_{o}^2+2gh##

    3. The attempt at a solution
    ##\sqrt{2gh}=4.43m/s##

    Would not one need to know the mass of the football to solve this problem? I find it uncomfortable and strange that the mass of the football is not included in this problem. Am I correct in this assertion?

    Thanks,
    KQ6UP
     
  2. jcsd
  3. Feb 7, 2017 #2

    TSny

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    Homework Helper
    Gold Member

    Yes, you need the mass in order to get a numerical value for the magnitude of the impulse. You might have to settle for specifying the answer in terms of the mass.
     
  4. Feb 7, 2017 #3

    jbriggs444

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    Science Advisor

    You can Google for the mass of a regulation football. (0.40 to 0.43 kg).
     
  5. Feb 8, 2017 #4
    Ok, I used .43kg, and I get:
    ##v=\sqrt{2gh}## implies that ##v=4.43m/s##.

    I used ##I=\Delta \boldsymbol{P}=\boldsymbol{P}-\boldsymbol{P}_o##

    Then componentized:

    ##\boldsymbol{P}_o=-1.90kg\cdot m/s \cdot \hat{y}##

    ##\boldsymbol{P}=18 m/s \cdot .43 kg \cdot (\cos 55^o \hat{y}+\sin 55^o \hat{x})##

    ##\therefore## ##\Delta \boldsymbol{P}## is a vector pointing 61.7##^o## with a magnitude of 9.36##kg \cdot m/s##. Is this correct?

    Thanks,
    KQ6UP
     
    Last edited: Feb 8, 2017
  6. Feb 8, 2017 #5

    TSny

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    Your answer looks correct. There appears to be a typographical error where your unit vectors ##\hat{x}## and ##\hat{y}## need to switch places in the expression for P.
     
  7. Feb 9, 2017 #6
    Yes, that is a typo. Thanks,

    KQ6UP
     
  8. Feb 9, 2017 #7
    Here it is fixed for the record.

     
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