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Vertical distance

  1. Oct 22, 2009 #1
    A kicker must kick the football from a point 29.4m from the goal and clear a bar 3.00m above the ground. The ball leaves the ground with a speed of 20.0m/s at an angle of 53.0 degrees above the horizontal.

    a) By how much does the ball clear, or fall short of the cross bar?

    I used the equation R=vi^2 sin2(theta)/g to see how far the football would travel.

    r+19.6 So the ball wouldn't even make it to the goal post. Correct?

    b) When it gets to the cross bar, what is the vertical component of the ball's velocity?

    This question makes me think my answer to a) is wrong because the x component would be zero. Correct?
     
  2. jcsd
  3. Oct 22, 2009 #2

    Delphi51

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    It does go over the bar. That range equation isn't much use here.
    I think you will have to write out the equations for the horizontal and vertical motion and find the time when the ball crosses the goal.
     
  4. Oct 22, 2009 #3
    When is the correct time to use that range equation?


    So, I used the equation
    yf=yi+vyit+.5ayt^2

    And found the time that the football is passing over the goal post to be t=3.06

    I then used the same equation and solved for yf using 3.00 as yi and the final y at that time is 6.0

    So, my answer to a) would be.....3.00m


    For b) I used the equation vyf=vyi+ayt

    My answer is -14.0

    How are those equations and answers?
     
  5. Oct 22, 2009 #4

    Delphi51

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    The range equation is supposed to give the horizontal distance it flies all right. I just meant it wouldn't help you find the answer to the question. My range equation is r = 2v^2sinθcosθ/g = 39 m.

    I don't agree with your time of 3.06 s. Did you work with the horizontal equation 29.4 = v*cosθ*t to get that?
     
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