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Soccer Ball Kick

  1. Aug 6, 2009 #1
    1. The problem statement, all variables and given/known data
    Your friend is standing on the roof of his house, which is 11.5 meters off the ground. You are standing some distance away from the house and you kick a soccer ball at a total initial velocity of 18.0 m/s @ 60 degrees above the horizontal from ground level. If the soccer ball gets to the top of the house at the peak of its motion, how far away were you standing?


    2. Relevant equations
    Tan(Θ)=Opposite/Adjacent


    3. The attempt at a solution
    I drew a triangle with the hypotenuse as the velocity vector and the side of the house as the side opposite the 60 degree angle so my distance from the house is the side adjacent to the 60 degree angle. Then, I used Tan(60) = 11.5/x and rearranged it to x = 11.5/Tan(60) and got 6.64 meters, which is wrong (the correct answer should be 14 meters). Could someone explain to me why my method was incorrect and explain how to get to the correct answer?
     
  2. jcsd
  3. Aug 6, 2009 #2
    That would work if your friend lived in space, but he doesn't.
    Consider how gravity factors into this projectile motion.

    Try and develop an expression of the form [tex]y(x)[/tex] describing the height of the ball off the ground as a function of its distance from you.
    Analyze the function to find what the relationship between the maximal height and its distance from the kick-off point is.
    Once you have that expression (I suggest you work parametrically up until this point), just substitute everything for the data in the question and you should get your answer.
     
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