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Projectile Motion - given angle, time; need displacement

  1. Jan 18, 2010 #1
    1. The problem statement, all variables and given/known data
    A ball is thrown at 60 degrees above x-axis. It remains in the air for 2.9 seconds. What is the final displacement in the x (range) if the ball was thrown with the same initial velocity, but at an angle of 45 degrees above x-axis instead?


    2. Relevant equations
    y = (Voy * t) + (1/2 * ay * t^2)
    Voy = Vo * sin(theta)
    Vox = Vo * cos(theta)
    x = 1/2(Vox + Vx)t


    3. The attempt at a solution

    So plugging into the first equation I get:
    y = (Voy * t) + (1/2 * ay * t^2)
    0 = (Voy * 2.9) + (1/2 * -9.8 * 2.9^2)
    Voy = 14.21

    Taking the result into the second equation I get the initial velocity:
    Voy = Vo * sin(theta)
    14.21 = Vo * sin(60)
    Vo = 16.41

    Am I on the right track so far? From here I would think I'd take it into the first equation again with the 45 degree angle so:
    y = (Voy * t) + (1/2 * ay * t^2)
    0 = (16.41sin(45)*t) + (1/2 * -9.8 * t^2)
    0=11.60t-4.9t^2
    t = 2.37
    So I've got the time, and Vox from Vox = Vo * cos(theta) since Vo is the same for both throws.

    For the equation
    x = 1/2(Vox + Vx)t
    (substitute Vox = Vo * cos(theta))
    x = 1/2(16.41cos(45) + Vx)2.37

    All I'm missing is Vx, I am lost where to go from here.
     
  2. jcsd
  3. Jan 18, 2010 #2

    ehild

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    Is there any horizontal force component?

    ehild
     
  4. Jan 18, 2010 #3
    No...
     
  5. Jan 18, 2010 #4

    ehild

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    No force, no acceleration. What does it mean for vx?

    ehild
     
  6. Jan 18, 2010 #5
    Is it Vx = Vo?

    Vx = Vo + 1/2*ax*t
    Vx = Vo + 1/2*0*t
    Vx = Vo
     
  7. Jan 18, 2010 #6

    ehild

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    Yes, you have got it !!!

    ehild
     
  8. Jan 18, 2010 #7
    I think I understand now, thanks a bunch. :>
     
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