# Motion of a projectile

We all are aware how we can use the forces and vectors to describe a classical projectile motion (No drag forces). Major things we already know include:

1. Position-time dependence of projectiles
2. Range of projectile
3. Distance traversed by projectile
4. Curvature at any point

My question relates to a 3D-Geometrical interpretation of the motion.

Consider a cubical environment with appropriate dimensions and align the x,y,z axes with the cube. We start a projectile at one of the bottom corner and throw it such that it reaches the diagonally opposite corner. Take the angle of launch to be $$\theta$$ and the initial velocity as $$u$$.

My interests and questions:

1. I want to find the system of equations of planes which will intersect at a point and give me the coordinates.
2. How many varieties of such planes are possible?
• I have some difficulty understanding your questions.First of all, how is the angle of launch measured? that is, which respect to which axis or other vector? And the coordinates of what, in your first question? Commented Aug 5, 2020 at 10:44
• @pglpm ...There are certainly many loop holes, i think I'll delete it, it doesn't look good. It was just a thought experiment to trace the trajectory of a projectile using an intersection of 3 planes Commented Aug 5, 2020 at 10:52
• Maybe it just needs some rephrasing rather than deletion. But consider that the very notion of coordinate is defined (in $\mathbf{R}^3$) in terms of intersection of planes. If you ask how many triples of non-coincident planes pass through a given point, the answer is "$\infty^{9}$" if I'm not mistaken; that is, they form a space with 9 dimensions. If you want them orthogonal then it's 4 dimensions. Check Grassmannians. So for each point of the trajectory there's a 9-dimensional space of plane triples passing through it. Commented Aug 5, 2020 at 11:00
• @pglpm this is informative, thanks, I'll check it out, I think I presented the easiest to find planes in my answer? Commented Aug 5, 2020 at 11:33
• You're welcome. Mmm... sorry I don't understand what you mean. What's easiest? I still don't quite understand which planes you have in mind... Commented Aug 5, 2020 at 11:39

That movement is ruled by an acceleration, which is the second derivative of position.

That means that you only need to specify two things: initial position, and initial velocity, (That includes magnitude and direction).

Given the initial position and the velocity, the plane in which the movement occurs is perfectly determined. So you would just choose that plane and work in two dimensions. It is not necessary to complicate it.

However, if you want to work in 3D, you can also do it. Just split your initial position and your initial velocity into its three components and you've got it:

$$\begin{cases} \displaystyle x=x_0+v_{ox}t+ \frac{1}{2} a_x t^2 \\ y=y_0+v_{oy}t+ \frac{1}{2} a_y t^2 \\ z=z_0+v_{oz}t+ \frac{1}{2} a_z t^2 \end{cases}$$

Edit:

Now that I see your comment, I understand what you mean.

Of course there are infinite possible moving planes whose intersection is one moving point.

The easiest solution is the one I said, just set:

$$\begin{cases} \displaystyle x=x_0+v_{ox}t+ \frac{1}{2} a_x t^2 \\ y=y_0+v_{oy}t+ \frac{1}{2} a_y t^2 \\ z=z_0+v_{oz}t+ \frac{1}{2} a_z t^2 \end{cases}$$

and those are actually three planes. The first one is $$x=x_{particle}$$, and so on.

Then, since they are planes

Any linear combination of those planes gives you another plane with the same solution.

That's the key of linear equations.

Check that all those planes depend on a parameter: $$t$$, which coincides with the time, but it is a free parameter after all.

One last thing, check that this question should belong to Mathematics SE

• I think the question asked to find equation of planes , you've described $r$ as a position of time Commented Aug 5, 2020 at 9:54
• You can use this as an addition to the community wiki i made Commented Aug 5, 2020 at 9:56
• You can always find the equation of a plane given two points and a vector... Commented Aug 5, 2020 at 10:41
• I think the question is very unclear, i should delete it... it's actually a thought experiment to trace the coordinates of a projectile using the intersection of 3 planes Commented Aug 5, 2020 at 10:53
• @AnindyaPrithvi you don't have to delete it, just edit it to make it clear. I don't understand what you mean with "coordiantes using intersection of 3 planes". Are you asking for the position of the projectile $\vec{r}(t)$ as intersection of 3 planes? Then choose the coordinates themselves: $x=x_p; y=y_p; z=z_p$ Commented Aug 5, 2020 at 14:58