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2 Dimensional Problem

  1. Feb 1, 2004 #1
    A quarter back throws a ball 18 m/s at an angle of 35 degrees above the horizontal. Standing 18 m away is the receiver. How far does the receiver have to go and at what speed must he travel to catch the ball.

    Assume that the delta Y is 0.
    Assume that the receiver leaves the same time the ball is thrown.
    No outside forces.
     
  2. jcsd
  3. Feb 1, 2004 #2
    What do u infer from \delta y =0

    Can u calculate the distance covered by the ball
     
    Last edited: Feb 1, 2004
  4. Feb 2, 2004 #3

    ShawnD

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    Science Advisor

    Neat question.

    First you have to find the time the ball spends in the air based on the Y velocity. Rearrange the distance formula (d = Vi*t + 1/2*a*t^2) or double the rearranged velocity formula (Vf = Vi + at). I personally prefer the velocity one since quadratics make things more difficult.

    [tex]t = 2(\frac{V_f - V_i}{a})[/tex]

    In this case, t is the total time in the air, Vf is 0, Vi is the initial velocity of the ball and a is gravity (remember that gravity is negative). Expanded looks like this

    [tex]t = 2(\frac{0 - 18sin(35)}{-9.81})[/tex]

    t = 2.1049

    Now find the distance traveled with d = vt.

    [tex]d = (18cos(35))(2.1049)[/tex]

    d = 31.036

    Now as for how far the guy has to run. He's already 18m away but he has to be 31.036m; he has to run 13.036m and he has to be there in 2.1049s. Just divide to get his speed.

    Double check my numbers though.
     
  5. Feb 2, 2004 #4
    Thanks that helps alot.
     
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