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Kicking a ball over a goal post

  1. Nov 18, 2012 #1
    1. The problem statement, all variables and given/known data
    This problem out of Halliday and Resnick is stumping me. A kicker can kick the ball with an initial velocity of 25 m/sec. The ball is 50 m from a 3.44 m high goal post.

    2. Relevant equations

    I have the range equation and the Path equations for a trajectory.

    3. The attempt at a solution

    The Range equation doesn't seem to help since it only will provide me ranges for various angles. I can easily write the parabolic path equation but I can't crack the math that gets me back to two angles (max and min ). y = (tan θ) X - gX^2/2(Vcos θ)^2
  2. jcsd
  3. Nov 18, 2012 #2
    What are you supposed to find? The range for the angle of a kick to go beyond the goal post?
  4. Nov 18, 2012 #3
    Sorry. We are asked to find the max and min angles between which the kicker must kick the ball in order to barely clear the goal post.
  5. Nov 18, 2012 #4
    "Barely clear" means the ball is just on top of the post. That gives you the end point of the trajectory. What are the relevant equations?
  6. Nov 18, 2012 #5
    The end point of the trajectory is the landing point(beyond the goal post) and may be determined with the range equation. But the question is not this--rather it is asking for the maximum and minimum angle the kicker must kick the ball in order to clear the post. We know the hight and distance to the goal post, and we know the initial kick velocity. I have the equation y = (tan θ) X - gX^2/2(Vcos θ)^2, but can't figure out how to solve it for the two angles. Or is there a simpler thing here I'm missing?
  7. Nov 18, 2012 #6
    You don't care about whatever happens after the ball clears the post. For the purpose of this problem, this point is the end point.

    That said, what you really need is an equation for the angle required to hit (x, y). This could be obtained from the trajectory equation you have, but it might be easier to start from the simpler equations of x and y as functions of time (and the angle/velocity parameters).
  8. Nov 18, 2012 #7
    Thanks but in order to get the time parameters one needs to know θ which is the unknown. Also, the two angles in question each give us a flight time that is different. I agree with you that working the path equation backward to find the θ's is hard, but I think this is the only way.
  9. Nov 18, 2012 #8
    The equations for x and y depend on time and the angle. You have two equations and two unknowns, so you should be able to find them both.

    If you really want to use the trajectory's equation, here is what you can do. Express the cosine via the tangent, or vice versa. Then let the cosine (or tangent) be the unknown variable. You will have an equation for that variable. Solve it. The do the arccos (or arctan) to get the angles. I recommend that you keep x, y and v in the symbolic form till the end, then plug the numbers in.
  10. Nov 18, 2012 #9
    We have D = VT or 50 = Tcos theta for the horizontal and 1/2gT2 sin theta for the vertical components, but while the horizontal motion equation could provide us with T, the vertical equation does not because the apogees of the two parabolic paths are not relevant to the question. I will work on your suggestion for the soln of the trajectory equation and thanks much. And even if I solve it and obtain an angle( say the lower angle for the kick) how do I find the larger angle that would also get the ball over the goal?
  11. Nov 18, 2012 #10
    The equation of the vertical displacement is a simple equation for uniformly accelerated motion; it would equally provide you with T. Then you could equate those T's from the h. and v. equations, and get an equation for the angle.

    The equation you will eventually get for the angle will be quadratic. There will be at most two solutions. There is no range of angles: there are either exactly one angle, or two, or none at all.
  12. Nov 18, 2012 #11
    Right. The vertical displacement is simply T x (V sin theta) - gT^2 and this works for so many problems of this sort, but not this one. P.S. tan and cos don't like eachother well enough to make this work so far as I can tell.
  13. Nov 18, 2012 #12
    Vertical displacement DOES work for this problem. In the end, the ball has to travel vertically to be just on top of the post. And it has to travel horizontally to be just there and then, too.

    Tan = sin/cos, and squared sin plus squared cos equals one. You can go from here.
  14. Nov 18, 2012 #13
    Thanks voko, I got it! Boy was I stuck on stupid. Thanks for being patient.
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