Understanding 3D Projectile Motion and Predicting Landing Spot

In summary: The angle between the velocity vector and \hat{i} is then:\theta_{0}=\arctan(\vec{V}_{0}\hat{i}-\vec{n})\
  • #1
Stanley_Smith
16
0
: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,
 
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  • #2
Stanley_Smith said:
...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.
 
  • #3
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]
 

FAQ: Understanding 3D Projectile Motion and Predicting Landing Spot

What is projectile motion?

Projectile motion is the motion of an object thrown or launched into the air, under the influence of gravity. It follows a curved path known as a parabola.

What factors affect projectile motion?

The factors that affect projectile motion are the initial velocity, angle of launch, air resistance, and gravity. These factors determine the shape and distance of the projectile's path.

What is the formula for calculating projectile motion?

The formula for calculating projectile motion is d = v0t + 1/2at2, where d is the distance traveled, v0 is the initial velocity, t is the time, and a is the acceleration due to gravity (9.8 m/s2 on Earth).

Can projectile motion be plotted on a graph?

Yes, projectile motion can be plotted on a graph with distance on the y-axis and time on the x-axis. The resulting graph will be a parabola.

Why is projectile motion important in science?

Projectile motion is important in science because it helps us understand the motion of objects in the real world, such as projectiles launched from a catapult or a ball thrown by a pitcher. It also has practical applications in fields like engineering and physics.

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