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Projectile Motion

  1. Sep 11, 2004 #1
    A ball is thrown from a bliff with an initial speed of 20m/s from the edge of a 45m high cliff. At the same time, a woman is running away from the cliff at a speed of 6m/s. She runs until she catches the ball. at what angle above the horizon should the ball be thrown so that she can catch ball.

    I came up with a position function for the woman

    [tex] V_{w} = 6t [/tex]

    and set it equal to the vertical position function for the ball.



    [tex] 6t= 45+ 20sin(X)t -\frac{-9.8}{2}t^2 [/tex]

    I cant see where I go from here. I tried using quadratic but I didn't think I could combine the 6t and the 20sin(x)t, am I incorrect? How do I do it with 2 variables, the x and the t?
     
  2. jcsd
  3. Sep 11, 2004 #2

    arildno

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    You are mixing up horizontal and vertical components of the ball's position.
    1. The ball reaches the ground when its vertical position is 0 (ground level)
    This gives you an equation for the time used, expressed in terms of the angle.
    2. At this time, the horizontal component of the position must equal the position of the woman.
    Air resistance is neglected.
     
  4. Sep 11, 2004 #3
    So . . .
    1. This would be ... [tex] 0= 45+ 20sin(X)t -\frac{9.8}{2}t^2 [/tex] right?

    2. [tex] 6t = 20cos(x)t [/tex] right?

    I understand that . . . but I don't see where to go. In equation 2, if I divide by 6t the t's cancel out and I get [tex] \frac{20cos(x)}{6} = 0 [/tex]

    Use quadratic to solve for the first equation and plug that into the 2nd?
     
  5. Sep 11, 2004 #4

    arildno

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    It should be:
    [tex] \frac{20cos(x)}{6} = 1[/tex], i.e [tex]\cos(x)=\frac{6}{20}[/tex]
    As it happened, it was unnecessary in this particular exercise to find the actual time value when the ball hits the ground.
    If the woman had started some distance off the cliff edge, you would have needed the particular time value.
     
    Last edited: Sep 11, 2004
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