**pre-reqs:** [kinematics](http://blog.dotphys.net/2008/09/basics-kinematics/)

My previous “basics” post was on kinematics (in one dimension). But what about two dimensions? In particular, what about projectile motion. My motivation here is that I was about to talk about analysis of a video that involved projectile motion and I don’t want to go over all the stuff again and again.

Let me start with a generic, one-dimensional kinematic equation:

![s kinematics](http://blog.dotphys.net/wp-content/uploads/2008/09/s-kinematics.jpg)

This relates the position (s), the velocity in the s-direction and the acceleration in the s-direction (oh, and time). What if I have a motion in 2-dimensions. I can write:

![kinematics](http://blog.dotphys.net/wp-content/uploads/2008/09/kinematics5.jpg)

If the x-variables (x, v_{x}, a_{x}) are independent of the y-variables (y, v_{y}, a_{y}) then the two kinematic equations can be treated independently. This leads to the key idea in projectile motion:

**Projectile motion is essentially two one dimensional motion problems that are independent of each other (except for time)**

The above holds true unless there is some odd situation like the x-acceleration depending on the y-velocity. In that case, you will not have a constant acceleration anyway.

There is a great demo you can do for yourself to show this idea. Take a book (or other object that can fall without destruction) and place it on the edge of a table so it is barely not falling. Then take some other object (you could use another book) and slide it towards the book on the edge. Hopefully what will happen is that both books will leave the table at the same time. One book will essentially fall straight down. The other book will be launched horizontally from the same height at the same time. Since both books start at the same vertical position and have the same initial vertical velocity (zero), they will hit the ground at the same time. The vertical motion is independent of how fast it is moving horizontally.

Here is a graph of such a thing. I have plotted the x-y position of three objects every 0.1 second:

![graph](http://blog.dotphys.net/wp-content/uploads/2008/09/graph.jpg)

First, look at the red dots. These are the dots that the book launched off the table with horizontal speed would make. As you have probably seen before, the trajectory is a parabola. Now, look at the blue dots. These represent the position every 0.1 second for a book that falls straight down from the same height at the same time. The important thing to look at here is that the y-position of both the red and the blue are the same. I added a third set of dots. Suppose there was a long table next to the edge of the other table. On this table I placed a block of ice that would slide at the same horizontal speed as the moving book. Note that these dots have the same horizontal position as the red dots.

But wait, what if I shot a bullet and dropped one from the same height? If you ignore effects of air resistance (and essentially even if you do) both bullets will hit the ground at the same time.

**Fine, but how do you SOLVE projectile motion problems?**

The basic strategy for projectile motion problems is to divide and conquer. You want to split the problem into 2 one dimensional problems. Use one of these problems to solve for the time and then use that value of time in the other 1-D problem. I will do one for you, because I like you. Suppose I shoot a ball horizontally off a table with a known height and known speed. How far will it go before it hits the ground?

![problem](http://blog.dotphys.net/wp-content/uploads/2008/09/problem.jpg)

So, what are the two 1-D problems? Assume that the acceleration in the vertical direction is -*g* and the acceleration in the horizontal direction is zero. Then:

![2 equations](http://blog.dotphys.net/wp-content/uploads/2008/09/2-equations.jpg)

If you are not sure where these come from, you should review the introduction to kinematic equations. Like I said before, the thing these two equations share is time. So, which one should I solve for time? The noob will say to start with the x-equation. But that is why they are noobs, that is a bad choice. Look at that equation. There are essentially two unknowns (we can choose the coordinate system such that x_{1} = 0 meters). The two unknowns are x_{2} and delta t. You can’t fully determine either of these with just one equation. But look at the y-equation. The only unknown is delta t. So, let me solve for delta t:

![delta t](http://blog.dotphys.net/wp-content/uploads/2008/09/delta-t.jpg)

Of course you could put some numbers in here, but really, that is beneath me (and it is boring). Now I will substitute this expression for delta t into the x-equation:

![x](http://blog.dotphys.net/wp-content/uploads/2008/09/x.jpg)

That is it. Note that I listed the horizontal velocity as just v_{x} without a 1 or 2 subscript. Since there is no acceleration in the x-direction, this velocity does not change.