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

[Stanley_Smith]

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

 

[arildno]

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.

 

[ShawnD]

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.

 

[Stanley_Smith]

"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,