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Projectile Motion football kick

  1. Oct 3, 2009 #1
    1. A place kicker must kick a football from a point 35.0 m (about 38 yd) from the goal. As a result of the kick, the ball must clear the crossbar, which is 3.05 m high. When kicked, the ball leaves the ground with a speed of 22.0 m/s at an angle of 52◦ to the horizontal. The acceleration of gravity is 9.81 m/s2 .

    a) To determine if the ball clears the cross-bar, what is its height with respect to the crossbar when it reaches the plane of the crossbar?

    b) To determine if the ball approaches the crossbar while still rising or while falling, what is its vertical velocity at the crossbar?

    2. Relevant equations

    3. The attempt at a solution

    No clue! I haven't done physics in three years and simply don't remember where to begin! Please help! (:
  2. jcsd
  3. Oct 3, 2009 #2
    First, write out the kinematic equations for each direction, both x and y. Since there is no acceleration in the x direction, you can solve for t (the time it takes for the ball to get the plane of the crossbar). You can then plug this into the equation for the y direction and solve for the height of the ball after a time t.
  4. Oct 3, 2009 #3
    Okay, so first I used x=Vxt to get a time of 1.6 s. Was that what I was supposed to do? Now, how do I find the height using that?
  5. Oct 3, 2009 #4
    That is what you were supposed to do. To find the height use the same equation you used for the x direction except, since the y direction has acceleration, you will have the extra term -(1/2)at^2.
  6. Oct 3, 2009 #5
    Okay. I have a question...as I was doing more questions, I noticed the equations Vxi=Vicostheta and Vyi=Visintheta.

    Was I supposed to use those for Vx because i just subbed in the velocity, not the x component of the velocity.
  7. Oct 3, 2009 #6
    Vxi = 22.0 m/s*cos(52)
    Vyi= 22.0 m/s*sin(52)
  8. Oct 3, 2009 #7
    So that gives me a time of 2.6 s instead.
  9. Oct 3, 2009 #8
  10. Oct 3, 2009 #9
    Now, using that same equation, just for y with the extraterm -(1/2)at^2, does that give me the velocity when it is over the bar? and is the 9.8 a negative value? If so ( or if not ) could you kindly explain why? How is height determined from this value? thank you very much!
  11. Oct 3, 2009 #10
    use y = v_o(t)+1/2a(t)^2. Use that to find the position of the ball after 2.6 seconds. Yes, 9.8 is negative due to the acceleration of gravity
  12. Oct 3, 2009 #11
    Okay, that gives me a velocityat 2.6 s of 11.9 if a is positive or 78.1 if a is negative....how do I know which to use?
  13. Oct 3, 2009 #12
    y = 22*sin(52)(2.6)+.5(-9.8)(2.6)^2

    That's how your equation should look.
  14. Oct 3, 2009 #13
    Okay, so y=11.9

    What is this y value? is it the velocity at 2.6 s?
  15. Oct 3, 2009 #14
    Yes, @ 2.6s.
  16. Oct 3, 2009 #15
    So would 11.9 be the answer for part b?
  17. Oct 3, 2009 #16
    subtract the height of the crossbar from y @ 2.6 seconds. When you do that, that will tell you by how much the ball clears
  18. Oct 3, 2009 #17
    Ah, I'm confused! subtract 3.05 from 11.9? what will that give me?
  19. Oct 3, 2009 #18
    11.9 - 3.05. = The height of the ball above the crossbar.
  20. Oct 4, 2009 #19
    I put that in for the answer and it says I'm wrong.
  21. Oct 4, 2009 #20
    Hoid on, let me check everything
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