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Simple projectile motion

  1. Dec 22, 2007 #1
    [SOLVED] simple projectile motion..

    1. The problem statement, all variables and given/known data
    A projectile is fired in such a way that its horizontal range is equal to 3 times its maximum height. What is the angle of projection?


    2. Relevant equations
    whole bunch for proj motion.


    3. The attempt at a solution
    I know that Dx = 3Dy. Let "t" be the time it takes for proj to reach max height.
    Dy = Vy*t (vertical)
    3Dy = 2Vx*t (horizontal)

    Vy/Vx = (2t*dy)/(3t*dy) = 2/3

    inverse tan (2/3) = 34 degrees, but the answer is 53.1 degrees. Any ideas what's wrong?
    Thanks.
     
  2. jcsd
  3. Dec 22, 2007 #2

    Doc Al

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    Staff: Mentor

    When Dy is max, Dx is only half the range.
    The vertical motion is accelerated, not constant speed.
    OK.
     
  4. Dec 22, 2007 #3
    Ok lol what a stupid mistake about the Dy. But now I'm not sure what to do... I know Vf = Vi + at. So Vy = 9.8t, and it still doesnt help me solve Vy/Vx cuz I get (9.8t^2)/(3Dy). Thank you for any help
     
  5. Dec 23, 2007 #4

    Doc Al

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    Staff: Mentor

    Hint: If Vy is the intial speed (in the vertical direction), find the average speed during the rise to maximum height. Dy will equal average speed x time.
     
  6. Dec 23, 2007 #5
    aha got it thank you for your help.
     
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