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Theoretical y position

  1. Sep 25, 2006 #1
    I have the initial X and Y velocity of a ping pong ball that was launched from a mechanical device at angle Theta (45 degrees). X_i = 3.378 m/s, Y_i = 2.027 m/s.

    I have a series of frames I've captured with a video capture program, which I pointed at the ping pong ball and the software calculated the X and Y position for that frame.

    I need to calculate the theoretical Y position for each frame using some formula....

    Which formula would accomplish this?
  2. jcsd
  3. Sep 25, 2006 #2


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    Homework Helper

    The y position is given with the equation of displacement of the ball in the y-direction, which is y(t) = y0 + v0*sin(Theta)*t - 1/2*g*t^2, where v0 is the initial velocity and y0 the initial y-coordinate.
  4. Sep 25, 2006 #3


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    Since a ping pong ball is very light, relative to it's volume, you must take into account air resistance.
    Drag is proportional to velocity squared and has a sense opposite to the velocity vector.
    [tex]D = kv^2[/tex]
    The parameter k is dependent of the cross section of the body, the viscosity and density of the air and of the form of the body. You must estimate it from the experimental data you have.
    Your equations of motion are:
    [tex]x(t) = x_0 + v_{0x}t -\frac{1}{2m}kv^2sin(\theta)t^2[/tex]
    [tex]v_x(t) = v_{0x} - \frac{kv^2sin(\theta)t}{m}[/tex]
    [tex]y(t) = y_0 + v_{0y}t -\frac{1}{2m}kv^2cos(\theta)t^2 - \frac{1}{2}gt^2[/tex]
    [tex]v_y(t) = v_{0y} - \frac{kv^2cos(\theta)t}{m} - gt[/tex]
    [tex]v^2 = v_x(t)^2 +v_y(t)^2[/tex]
    [tex]\theta = tan^{-1}\frac{v_y(t)} {v_x(t)}[/tex]
    These equations must be solved numerically.
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