Teaching Script 
The following is a script I roughly followed in
giving my presentation. There have been changes to the theory and
actual flow of the presentation, but this was my origninal idea.
What I am going to talk about today is Vector Kinematics in Two Dimensions… or in English, How do projectiles move in the air.
What is considered a projectile? As I toss this object in the air it becomes a projectile. So a projectile is something moving through the air given some initial push or toss, but after that has no means to keep itself in the air on it’s own. So a bird or a rocket would not be a projectile in this sense.
We would like to know how to describe the motion of a projectile. We would ask such questions as what affects the projectile path while in the air? What path does it take in the air? How long will it stay in the air? How high will it go and how far? In real life we could use projectile motion analysis to figure out if I'd be able to throw a ball over a fence, or how far I could hit a golf ball. One question I would like to ask you to think about while I talk about the projectile motion today if I were to jump from a car moving at constant velocity, in what direction should I jump if I want to land back in the car? At the end I will use a demonstration to illustrate the answer.
Before we start to get equations that will describe projectile motion we make a couple of assumptions and observations to make our analysis easier. First off, although we are in a 3D world all projectiles can be treated in 2D if we just rotate ourselves so we are looking the projectile simply moving vertically and horizontally and not moving closer or farther away from us. Next we have to think about gravity. The effect or gravity lessens as we move further from the earth, but if we assume that the displacement of the projectile is much smaller than the radius of the earth, we can assume that gravity is constant at 9.8 m/s2. Another assumption that we will make is that air resistance is negligible. This can be done for objects where the surface area to mass ratio of the projectile is small.
With these assumptions in mind we can go about analyzing projectile motion.
One more fact that makes our analysis of projectile motion easier is the fact that the x and y components of a projectile motion are independent of one another. Meaning we can treat the motion of the particle moving horizontally and vertically separately.
From one dimensional vector analysis we know that general formula for velocity at a given time given an initial velocity is:
If we look at the horizontal component we ask
ourselves:
What affects the projectile in the horizontal
direction? The answer to this is nothing!
And so the acceleration term can be cancelled
out and we are left with:
From this equation we can see that the horizontal velocity of a projectile is constant under the assumptions we made earlier.
How about the vertical component? Does anything affect it motion in the vertical direction? Yes, gravity… and thus our equation becomes:
Notice the sign before the g is negative. This happens because if we choose the positive vertical direction as up gravity acting downward must be negative.
The position of the projectile can be treated much in the same way:
Given the general formula for position from 1D vector analysis:
We can get the position in x from as:
And the position in y as:
Notice I’ve removed the x0 and y0 terms since we will be starting from x=0 and y=0
The above four equations would be all we need if we were always given the velocity of the projectile in it's x and y components. Unfortunately, more often than not, this is not the case. What we usually know or are given are an initial velocity and some initial angle from which it left the ground.
This adds another step to our analysis. Namely it requires us to resolve the initial velocity into its x and y components using trigonometry.
Now we can use these resolved components and replace the vxo and vyo in our equations:
<<overhead 3: equations for velocity and position with angle>>
Now we have equations that can help us describe where our projectile is in time given a particular initial velocity and angle.
We start answering some of the questions asked earlier if we notice that we can solve for t using the equation for x.
<<overhead 4: solving for t with equation for horizontal position>>
If we substitute this into the equation for y and rearrange the terms a bit we can see that the equation looks like a quadratic formula.
And the quadratic formula has a particular shape, a parabola.
And now we can get more information namely the peak height and range of a projectile. Both of these properties can be found by first figuring out how long it will be or was in the air. One can observe that a projectile will be at y=0 at two times. Once when it first leaves the ground and once at the point when it lands. The first situation will be at t=0 and the second situation will be when t=the time at spent in the air. So setting y=0 in our equation and solving we can get t:
<<overhead 5: solving for t with vertical position equation>>
If we substitute this value of t into our x equation we get:
Which is the range of the projectile. Notice that if initial velocity is held constant the range will only vary with the angle. So this range will be a maximum when sine of two theta is a maximum. Since sine goes between 1 and –1 we want sine of two theta to equal 1 and this is when theta is 45 degrees. So one can achieve maximum range by shooting a projectile at 45 degrees. Other angles above or below 45 degrees will result in lesser ranges. Also note that because of this fact, I can obtain the same range with two different angles:
For the peak we can see from the parabola, this occurs at halfway so at total time divided by two:
If we substitute this time into the equation for y and after some algebra we get:
And now I’d like to show you this device, which is a water projector that will shoot a stream of water at some fixed initial velocity and a given angle which I can vary. You can consider it as shooting out a series of projectiles one after another, each one following projectile behaviour.
As you can see that the path the water stream takes is in fact a parabola.
As I vary the range you can see that the range will indeed change. With a 45 degree angle resulting in a maximum range. As I increase the angle the range will be reduce and I can also obtain the same range if I decrease the angle. Thus proving I can get the same range from two different angles.
Now for my final demonstration I would like
to go back to my question I asked earlier if I had to jump from a car and
I wanted to land back in the car in what direction should I jump.
(pole the class)
If you can remember, one of the first things
I said we can do to analyse projectile motion was we can use the fact that
the x and y components are independent. If this is the case it shouldn’t
matter how high I jump if I’m in the car initially I’m moving at the same
velocity of the car. When I jump, if air resistance is ignored, my
x component of velocity should remain constant with that of the car also
moving at constant velocity. If I jumped exactly vertically upward,
I should land back down exactly from where I jumped. Again using
a water stream as a series of projectiles you can see as I accelerate the
car the stream will move in an arc, but as the car approaches constant
velocity the stream will maintain a vertical path.
