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Projectile motion of a football

  1. Sep 12, 2007 #1
    The kicker on a football team can give the ball an initial speed of 26.2 m/s. If he is to score a field goal from a point 42.2 m in front of goalposts whose horizontal bar is 3.75 m above the ground, what is the minimum and maximum angle above the horizontal he must kick the ball?

    I figured out this much

    ax=0
    ay= -9.8
    vx= sin(theta)*26.2
    vy= cos(theta)*26.2
    x= sin(theta)*26.2x
    y= cos(theta)*26.2x-4.9x^2

    and I think

    3.75= cos(theta)*26.2x-4.9x^2

    My first attempt at an answer yielded 5.06 and 26.something. I tried using arctan(3.75/42.2) to get an angle. I'm not really sure how to approach this problem since theta is missing.
     
    Last edited: Sep 12, 2007
  2. jcsd
  3. Sep 12, 2007 #2

    learningphysics

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    By 'x', do you mean time? I'm a little confused by 'x' when it is used for multiplication along with as a variable...
     
  4. Sep 12, 2007 #3
    yes, x for time
     
  5. Sep 12, 2007 #4

    learningphysics

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    You have the vx, vy mixed...

    It should be:
    vx= cos(theta)*26.2
    vy= sin(theta)*26.2

    So the two equation become (I'm going to switch to t becomes we have another x already :wink:):

    42.2= cos(theta)*26.2t
    3.75= sin(theta)*26.2t-4.9t^2

    solve these to get theta.

    Then you also should solve:
    42.2= cos(theta)*26.2t
    0= sin(theta)*26.2t-4.9t^2

    For this case, the second equation simplifies to:
    0= sin(theta)*26.2-4.9t

    by dividing both sides by t.
     
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