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Projectile motion in 3-D

  1. May 14, 2004 #1
    :mad:

    Hi Everybody,

    I'm currently involved in a project in which I have to display the trajectory of a flying ball in 3D and predict its landing spot. My partners will track the ball as it is launched and give me a set of the ball's 3-D coordinates. The display path is easy but I have a few questions about the predicting path:

    Normally, the object's landing spot in 2-D will be calculated by the following formula: (v^2*sin(2theta))/g
    where v is the initial velocity, theta is the launching angle and g is gravity

    Now, I never been exposed to projectile motion in 3-D and I have a few questions:
    How do I extract the launching angle from a set of 3-D coordinates ?
    And I am thinking about using the 2-D equation above to calculate where the ball will land (in 2-D) and then somehow obtain the third dimension in the end...Is this a right approach ?

    Thank you very much,
    Stan
     
  2. jcsd
  3. May 14, 2004 #2

    arildno

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    Welcome to PF!

    "And I am thinking about using the 2-D equation above to calculate where the ball will land (in 2-D) and then somehow obtain the third dimension in the end...Is this a right approach ?"

    This is a very good approach, because the the trajectory, [tex]\vec{r}(t)[/tex], will lie in a plane whose vector normal is proportional to the cross product of the initial velocity vector, [tex]\vec{v}_{0}[/tex], and the constant acceleration vector, [tex]\vec{a}[/tex], i.e [tex]\vec{v}_{0}\times\vec{a}[/tex]

    Hence, the trajectory curve is in essence a 2-D curve (its torsion zero).

    As for expressing the launching angle, the closest analogy to the 2-D case is the polar (azimuthal??) angle in spherical coordinates.
     
  4. May 14, 2004 #3

    ShawnD

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    You can probably just cheat by drawing a line between where you launch it and where it lands, then draw a perpendicular line along the ground and mark it as your Z axis.
     
  5. May 20, 2004 #4
    "because the the trajectory, [tex]\vec{r}(t)[/tex], will lie in a plane whose vector normal is proportional to the cross product of the initial velocity vector, [tex]\vec{v}_{0}[/tex], and the constant acceleration vector, [tex]\vec{a}[/tex], i.e [tex]\vec{v}_{0}\times\vec{a}[/tex]

    Hence, the trajectory curve is in essence a 2-D curve (its torsion zero)"

    Arildno, could you please explain more about the cross product stated above?
    And What do you mean by "will lie in a plane whose vector normal is proportional "
    I kinda understand what you are saying, but I'm not sure...

    Thank you,
     
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