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Projectile motion theta question

  1. Oct 11, 2005 #1
    A projectile is launched from origin at angle theta with horizontal, whose position is given by r(t). For small angles, distance from origin always increases. But if projectile is launched nearly straight up, it goes to a farthest point and then moves back down towards origin, so distance to origin first increases, then decreases. Which initial launch angle theta divides the two types of motion?


    I set up an equation of x^2+y^2 where x = vcostheta and y = vsintheta - 4.9x^2 and then took the derivative of it with respect to t, setting it equal to 0. However, that gets really messy and is there a better way to do it? I'm not sure about my answer anyways. Can someone please guide me though the process? Thanks.
  2. jcsd
  3. Oct 11, 2005 #2


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    I did manage to get an answer by doing the same things as you did (derivating and setting that 0).

    It was surprisingly short, even though it seemed quite long at the beginning. I got 71 degrees, which seems somewhat logical (I'm telling this as you don't seem to have the correct answers in the book).
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