Projectile Motion in 3D: Calculating Landing Spot and Launching Angle

AI Thread Summary
The discussion focuses on predicting the landing spot of a projectile in 3D space using initial velocity and launching angle. The 2D landing spot formula, (v^2*sin(2theta))/g, is deemed a good starting point, with the trajectory being essentially a 2D curve within a 3D plane. To extract the launching angle from 3D coordinates, the trajectory can be analyzed using spherical coordinates, specifically the polar angle. The cross product of the initial velocity vector and the acceleration vector helps define the plane of motion. Additional clarification is sought on the cross product and its implications for the trajectory's orientation.
Stanley_Smith
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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
 
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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, \vec{r}(t), will lie in a plane whose vector normal is proportional to the cross product of the initial velocity vector, \vec{v}_{0}, and the constant acceleration vector, \vec{a}, i.e \vec{v}_{0}\times\vec{a}

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

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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