## Constant Acceleration in Two Dimensions

### Motion on an Incline Plane

Something sliding down an incline plane can seem like it is moving in two dimensions because it is not perfectly parallel or perpendicular to the surface of the earth. That is only because we typically arrange our coordinate systems with that orientation! We can orient our coordinate systems any way we would like! If motion is confined to one dimension (here along the plane) it is often good to make one axis along that direction.

In the video, we look at that and provide a tutorial on how to deal with the geometry and trigonometry of wedges, inclines, projections, and angles that will come in handy throughout the course.

### Constant Acceleration in Two Dimensions

Now we want to look at 2D motion with constant acceleration. There is not too much different between this and solving similar problems in 1D, except you have to keep track of the parameters in both dimensions. The bookkeeping can get extensive. Lots of space and be clear and careful with your notation.

The key is that motion in one dimension does not affect the motion in the other dimension. Thus, working in 2D is like working in 1D twice, one in each dimension. There are not two versions of time, however, and that is the link between the two dimensions. Each point in time, the object is at a single set of coordinates. It is clearest to see how this works through examples.

### Constant Acceleration in 2D: Example

Since this is best taught through example, let's do another one. A rock is thrown horizontally with speed \(v\) on a wedge that makes an angle \(\phi\) with the horizontal. How far down the incline does it get when it hits?

### Projectile Motion without Time

For two-dimensional motion it has been stressed that the motion in one dimension does not affect the motion in the other dimension. The connection between them is time, since the particle can only be in one place at one point in time. We can use that by solving for time in one dimension and substituting it in the other to establish a relationship between the two coordinates.

Here we do this for the special case of projectile motion, not too far, near the surface of the earth. An object is launched with a speed, \(v_o\) at an angle \(\alpha_o\) with respect to the horizontal. In deriving this equation, we fix the coordinate system to have +x be parallel to the surface of the Earth in the direction the object is launched and fix the +y direction to be up. The origin of the coordinate system is also fixed to be at the launch point. It is important to remember if using this equation, that you do **not** have the freedom to choose your own coordinate system. The choice of coordinate system is already baked into the equation.

With those caveats, the parabolic trajectory of projectile motion is given by: $$y(x) = (\tan \alpha_o)x - \frac{g}{2v_o^2 \cos^2\alpha_o}x^2 $$

### The Range Equation

For an even more constrained set of circumstances we can find an even simpler equation. Take everything from the projectile motion topic above and add that the final vertical position is the same as the initial. This comes from asking the question: if you launch something from the ground (initial speed \(v_o\) and initial angle \(\alpha_o\) from horizontal), how far does it go when it hits the ground? You get the idea where the name comes from.

Given those assumptions and the ones from the previous topic, including that the origin of the coordinate system is at the launch point, the final x coordinate when the object returns to its y = 0 is given by: $$ x = \frac{v_o^2}{g}\sin (2\alpha_o) $$

### Exercises

*Do Now!* Do these exercises immediately.

*Not Now!* Do these after you start to forget the topic, say in a week.

*More!* More exercises if you want. Maybe review before a test.

### Required Problems

### Solution

1. You launch a ball with a speed of 24 m/s a direction of 30 degrees above the horizontal. At the moment of launch, it is 1/2 meter above the floor. It hits a wall 40 m away. How far above the floor is it when it hits the wall?

### Solution

2. You are trying to jump over a ravine on your motorcycle. The ravine has a width of \(W\) and the other side is \(h\) meters higher than your level. You launch with a speed \(v\) and angle \(\theta\). For a given angle, what is the maximum \(h\) can be such that the jump is possible? For any possible jump, what is the minimum speed such that you just make the jump. If \(v = 30\) m/s, \(\theta = 60\) degrees, \(h = 30\) m, \(W = 100\) m, and ravine depth is 100 m where do you land?

### Solution

3. An ion is traveling very quickly with speed \(v\), when it enters half-way between two deflector plates separate by width, \(w\) and enters parallel to the plates. The plates have a length, \(l\) and while between the plates, the particle experiences an acceleration \(a\) perpendicular to the plates. Once outside the plates, the ion is a distance, \(d\) from a detector, where \(d >> l > w\). How much was the particle deflected once it reaches the detector?