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Projectile motion again

  1. May 23, 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 ?

    There was an answer from Arildno, Thanks Arildno:

    "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 will lie in a plane whose vector normal is proportional to the cross product of the initial velocity vector and the constant acceleration vector

    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.

    But:

    What does he means by :

    "because the trajectory will lie in a plane whose vector normal is proportional to the cross product of the initial velocity vector and the constant acceleration vector"

    I don't think I understand what he is saying? Can anyone help ???

    Thanks a lot,
     
  2. jcsd
  3. May 23, 2004 #2

    Doc Al

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    Staff: Mentor

    Regardless of the initial velocity, the motion will be in a 2-D plane. arildno is telling you how to find that plane. Another way to find the plane: find the projection of the initial velocity on the x-y plane. The plane of the trajectory will intersect the x-y plane at that same angle. The reason is that gravity only affects the z-component of motion; the x and y velocity components remain constant.
     
  4. May 23, 2004 #3

    arildno

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    Dearly Missed

    Sorry for not making myself clearer, Stanley!

    1.Suppose you have an initial velocity vector:
    [tex]\vec{V}_{0}=V_{x,0}\vec{i}+V_{y,0}\vec{i}+V_{z,0}\vec{k}[/tex]
    The constant acceleration of gravity vector is given by:
    [tex]\vec{g}=-g\vec{k}[/tex]

    Hence, the vector unit normal to the plane in which the trajectory lies is proportional to:
    [tex]\vec{g}\times\vec{V}_{0}=-gV_{y,0}\vec{i}+gV_{x,0}\vec{j}[/tex]

    2.Requiring that the unit normal has unit length, we find:
    [tex]\vec{n}=\frac{V_{y,0}\vec{i}-V_{x,0}\vec{j}}{\sqrt{V_{y,0}^{2}+V_{x,0}^{2}}}[/tex]

    3.The origin is in the plane, so the equation for the plane of trajectory becomes:
    [tex]\vec{n}\cdot\vec{x}=0\rightarrow{V}_{y,0}x-V_{x,0}y=0[/tex]

    4. We will now express the launching angle as the angle between the velocity vector
    and a natural choice of unit vector in the given plane, normal to the vertical.
    We will call that unit vector [tex]\hat{i}[/tex]

    Clearly, the initial velocity vector may be written:
    [tex]\vec{V}_{0}=V_{0}(\cos\theta_{0}\hat{i}+\sin\theta_{0}\vec{k})[/tex]
    where:
    [tex]V_{0}=\sqrt{V_{x,0}^{2}+V_{y,0}^{2}+V_{z,0}^{2}},[/tex]
    [tex]\hat{i}=\frac{V_{x,0}\vec{i}+V_{y,0}\vec{j}}{\sqrt{V_{x,0}^{2}+V_{y,0}^{2}}},[/tex]
    [tex]\cos\theta_{0}=\frac{\sqrt{V_{x,0}^{2}+V_{y,0}^{2}}}{V_{0}},\sin\theta_{0}=\frac{V_{z,0}}{V_{0}}[/tex]
     
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