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2d Kinematics

  1. Nov 11, 2008 #1
    1. The problem statement, all variables and given/known data
    A rocket is launched at an angle of 75 degrees and accelerates at 21.4 m/s^2. Where is the rocket located (from the point of origin) after 11.2 seconds?

    2. Relevant equations
    a=(vf - vi) / t

    (vf + vi) / 2 = (xf - xi) / t

    where: vf = velocity final, vi = velocity initial, xf = displacement final, xi = displacement initial, and of course a = acceleration and t = time

    3. The attempt at a solution
    given solution: Xfx = 353.7 m; Xfy = 693.4 m (don’t forget gravity in the y-direction)

    ax = 21.4cos75 = 5.54
    vix = vi cos75
    vfx= vi cos75
    xfx = ?
    xix = 0
    t = 11.2

    ay = 21.4sin75 = 20.67
    viy = vi sin75
    viy = ?
    xfy = ?
    xiy = 0
    t = 11.2

    any one got any Ideas to the path to the solution, and how do i account for gravity? add/subtract?

  2. jcsd
  3. Nov 12, 2008 #2
    I believe your main problem is that you don't have the equation

    d = vt + (1/2)at^2 ; where v is initial velocity, t is time, and a is your acceleration.

    Once you break the acceleration into it's x and y components, you should be able to get both the x and y distances with the above formula.

    Also one thing to note is that the acceleration in the x direction doesn't get effected by gravity, and the acceleration in the y direction is the difference between your original acceleration and gravity. (i.e initial velocity - gravity)
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