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Football related Projectile problem!

  1. Jan 25, 2008 #1
    1. The problem statement, all variables and given/known data
    A placekicker must kick a football from a point 36.0 m (about 40 yards) from the goal, and half the crowd hopes the ball will clear the crossbar, which is 3.05 m high. When kicked, the ball leaves the ground with a speed of 20.0 m/s at an angle of 48.0° to the horizontal.
    (a) By how much does the ball clear or fall short of clearing or fall short of clearing the crossbar? (Enter a negative answer if it falls short.)


    2. Relevant equations
    Possibly:
    vxi = vi cos theta
    vyi = vi sin theta
    change(y) = viy(t) + .5(a)(t)^2


    3. The attempt at a solution
    20 sin 48(t) - 4.9(t)^2
    = 26.75-4.9(1.8)^2
    change(y) = 26.75 - 15.88 =10.87

    10.87-3.05 = 7.82 m. Which is the wrong answer.

    Can someone please help. Thanks
     
  2. jcsd
  3. Jan 25, 2008 #2

    Doc Al

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    Staff: Mentor

    Start by figuring out the time it takes for the ball to reach the goal. Hint: Consider the horizontal motion.
     
  4. Jan 25, 2008 #3
    time = (vi sin theta) / g

    Is that the correct equation?

    So it's (20 sin 48) / 9.8 = 1.52

    I plugged that in and obtained 8.21 m as my finals answer, which is incorrect. Can some please tell me what I did wrong?
     
  5. Jan 25, 2008 #4

    Doc Al

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    Staff: Mentor

    No, that's not correct. Hint: The horizontal component of motion is unaccelerated. (Gravity only acts downward.)

    What's the speed in the horizontal direction?
     
  6. Jan 25, 2008 #5
    (20 cos 48) / 9.8?
     
  7. Jan 25, 2008 #6

    Doc Al

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    Staff: Mentor

    Why divide by 9.8? (Check your units.)
     
  8. Jan 25, 2008 #7
    Opps, I thought I was still looking for time.

    Why would I need to know the speed though, since the velocity is already given.

    I basically just plug everything into the R = (vi^2 sin 2theta)/g equation right?
     
  9. Jan 25, 2008 #8

    Doc Al

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    Staff: Mentor

    That's what we're heading towards calculating.

    Then when I asked for the speed in the horizontal direction you should have answered without hesitation. :wink:

    Nope. That's a range equation.

    Start with what I suggested in post #2.
     
  10. Jan 25, 2008 #9
    The Range equation only applies to situations when your initial and final y values are the same. In other words, y is unchanged.
     
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