projectile motion again...

[Stanley_Smith]

: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,

[Doc Al]

...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"

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.

[arildno]

Sorry for not making myself clearer, Stanley!

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

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

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

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

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 \hat{i}

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