In this chapter we'll start by looking at trigonometry. The results of this are listed below. We'll then look at the 2-dimensional addition of displacement, velocity and force vectors. We'll finish up with a look at projectile motion.

1. Trigonometry

For plane geometry, as opposed to the geometry of curved surfaces, there are certain relationships among the sizes of angles and the lengths of the sides of triangles that are always true. (In the figure, the blue triangle, shares part of two legs with the similar red triangle.)

1. For any triangle, the sum of the interior angles is equal to 180º.

In this triangle,
α + β + γ = α + β + γ = 180º

For those of you who don't speak Greek, the angles above are pronounced "alpha", "beta", and "gamma" respectively.

2. The ratio of the lengths of any two sides of any triangle will be equal to the ratio of the lengths of the corresponding sides of any other similar triangle. Similar triangles are triangles with the same interior angles.

For example, for the triangle shown,

x/y = x/y

Note: To indicate two corresponding values like x & x we'll generally use unprimed & primed values (x & x'). The colors are just nice in this video environment.

We will be primarily interested in right triangles, that is, triangles with a 90º angle. For these triangles, all of the above is still true, but we can more specifically state the following.

1. The lengths of the sides of of a right triangle are related by the Pythagorean Theorem which states: c2 = a2 + b2

For the triangle shown,

• a is the side opposite θ
• b is the side which is adjacent to θ, but not opposite the 90º angle
• c is the side opposite the 90º angle. It is the longest side, and it is referred to as the hypotenuse.

2. For a given right triangle and any triangle similar to it, the ratios of the lengths of the sides, will always be the same. For our triangle,

a/c = a/c

b/c = b/c

a/b = a/b

For any right triangle with some angle θ as one of its acute angles, each of these ratios will be the same. We have tables of these ratios in books and in our calculators. The three ratios we're interested in have the names sine, cosine, and tangent. They are defined as follows.

Trigonometric Functions sin θ = a/c = opposite/hypotenuse cos θ = b/c = adjacent/hypotenuse tan θ = a/b = opposite/adjacent

We will do much more with trigonometry in class, but that's it for the basics.

2. Projectile Motion - Horizontal Launch

Projectile An object moving under the influence of gravity alone.

There are zillions of examples of projectiles around us. And we know a lot about them at some physical level. With a little practice we can catch a fly ball, or shoot an arrow at the bulls eye. With no practice we can jump across a creek and land on just the spot we want.

Some projectiles are weirder than others.

Sorry.

The simplest case of a projectile is a dropped object. Piece a' cake. We need to look at the 2-D version now. Our goal is to be able to predict from initial velocity information (speed & direction), the velocity & location of an object at any later time. If a ball is kicked at a 30º, angle at 10 m/s, where will it land or will it go over the goalie's head?

We'll start with the simplest case, a ball launched with an initial horizontal velocity. Here it is in 3-D. Run this movie a few times. Look at the shadows and the grid lines. Got any ideas of how we're going to tackle this motion? (For a big hulkin' version, try this.)

There is a lot of good information there if you know how to look for it. But you need to discover it yourself. Let's go to another movie of a similar motion and collect some data.

Here's the motion we want to study. While moving at a constant, horizontal velocity our low friction cart drops a ball. Here's the apparatus. The curved rod has a magnet that holds the ball. A photogate tells the electromagnet when to release. Here's the motion slowed down to 1/4 speed. Watch it a few times. Step through the frames. Make some observations.

The actual motion is quite complex. It's a parabola. We could deal with this shape mathematically, but there's a better way. We're going to divide and conquer. That is, we'll look at the horizontal motion alone, then at the vertical motion alone. Look again at that last movie. Do you see where we're going now?

Horizontal Motion

Let's get some data with Frame Flipper. Here's what you need to do.

1. Collect data in Frame Flipper.
2. Copy time, t, horizontal position, x, and vertical position, y data into a table in GA. (You'll need to add a column.)
3. Create a new column with v_vertical
   (V_vertical is the slope of your y,t graph, a.k.a. the derivative of y in GA. Review Acceleration 3b if you need help going from position to velocity in GA)
4. Plot x,t.
5. Plot y,t.
6. Plot v_vertical,t
7. Print out your three graphs, then come back here for discussion. (Don't close GA.)

1. Look at your x,t graph. What does it represent? This is a plot of the horizontal position of the ball vs. time. It completely disregards the vertical motion.

What does your graph tell you?

velocity in x-direction is constant.

acceleration in x-direction is constant (and non-zero).

What does it mean that the horizontal component of the velocity is constant? Let's revisit a movie that should clarify this observation. The cart moves at a constant horizontal velocity. The rod that holds the ball is attached to the cart, so it moves at the same speed. You should see that the ball stays right with the rod, so the ball keeps going at the same speed as the cart, that is, the speed it was going when it was released.

Try it yourself. Get a ball or an eraser or whatever. Find a safe place and run along with the ball then drop it beside you. You'll see that it hits the floor right beside you.

A projectile moving in two dimensions moves horizontally at a constant velocity.

We haven't really proved this, but it follows nicely from Newton's 2nd Law. Since there is no net force in the horizontal (assuming no friction) there should be no horizontal acceleration.

Let's use this new knowledge. You just found the horizontal velocity, v_x. Using that information, where should the ball be after the first .33s of the animation? Well, we don't know its altitude, but we can calculate how far it travels horizontally during this time.

Δx = v_xt = 1.17 m/s x .33 s = .39 m

Let's check it and see. Here's the same movie at full speed. Notice the ball's initial horizontal position. Then step through the frames of the movie. Each one is 1/30 s. Since there are 10 frames, the total time is 1/3 second. Notice the final horizontal position.

Everyone with me? We've established that if we know the horizontal velocity and how long it's been moving, we can get the horizontal displacement. Let's proceed to the vertical motion.

4. A ball is thrown horizontally at 20 m/s from atop a tall building. How far has it traveled horizontally after 3 seconds?

Vertical Motion

There are a couple of good ways to find out. Let's go back to our first animation. Step through it a few times, looking at the shadows cast on the two walls and the floor. The left wall indicates the real parabolic motion of the ball. The grid lines on each wall are drawn to show where the ball (actually its shadow) is every, say, 1 second.

The floor shows the horizontal component of the motion. Notice the spacing of these grid lines. Their equal spacing indicates the constant horizontal velocity.

5. Now look at the spacing on the right wall which indicates the vertical component of the motion. What does this spacing seem to indicate about the vertical motion?

	velocity in y-direction is constant.
	acceleration in y-direction is constant.

Well, it's at least speeding up, but how can we be sure that it has a constant acceleration? We need data. Let's get back to our GA graphs.

6. Look at your y, t graph. What about this graph indicates that the ball is speeding up?

So it is accelerating, but is it a constant acceleration? Look at your vertical velocity, time graph. Here we see proof of our constant acceleration.

7. What is that proof and what is the value of this acceleration?

Let's add this to what we know about horizontal motion.

Projectile Motion A projectile moving in two dimensions moves horizontally at a constant velocity. Its vertical motion is a constant acceleration at g.

It's just that simple.

Let's return to our initial movie of the pink ball. We found that it traveled about .39 m during the first 1/3 second. How far should it fall vertically? You figure it out; it's just free-fall. Do the calculations then go back and see how you did. This image shows the ball at the beginning and end of this interval. You should be able to use the metersticks to check your calculations.

Let's be sure we have this straight. We've found that the parabolic motion of a projectile can be divided up into its two component motions.

1. It moves horizontally at a constant velocity.

2. It moves vertically at a constant acceleration. (We'll add some initial vertical velocity later.)

Here are the calculations for the problem we just worked.

1. We were given a time interval of 1/3 second and a vertical acceleration of -9.8 m/s².

2. The horizontal distance it traveled was given by:
Δx = v_xt = 1.17 m/s x .33 s = .39 m

3. The vertical distance it traveled was given by:
Δy = 1/2 a t²
 = 1/2 (-9.8 m/s²) (.33 s)² = -.54 m

Our movie concurred so we seem to be on good footing here. We'll get plenty of practice with this technique in class, but let's elaborate a bit first.

Knowing that we have constant horizontal velocity (a_x = 0) and constant vertical acceleration, we can write the 5 kinematics equations for each dimension as follows.

Horizontal	Vertical
x = v(const)xt	y = v(avg)yt
	v(avg)y = (voy + vfy)/2
	ay = g = (vfy - voy)/t
	vfy2 = voy2 + 2ayy
	y = voyt + (1/2)ayt2

The first column has only one entry since a_x is zero. Can you see why?

There are a lot of problems that can be solved using these equations and the following observation.

8. What is the common variable that occurs in both the x & y dimensions for projectile motion?

x

vx

t

g

y

Here's a typical problem.

A stone is thrown horizontally from a 20-m-high cliff at 4 m/s. How far from the base of the cliff will it hit?

Horizontal:
We know v_x & we want x. We need t.

We can find t from the vertical.
Vertical:
We know a = -g and y = -20 m. We can find t from equation 5.

Plugging t into x = v_x t will give us x.

Try it. You should get x = 8.1 m (using g = -9.8)

3. Projectile Motion - Non-Horizontal Launch

Here's our mathematical method in a nutshell. If an object is launched at some angle with the horizontal, it will start with both a horizontal and vertical component of velocity. We'll use these values in our kinematics equations. The rest is the same as in the horizontal launch.

These velocity components are found from:

Velocity Components vox = vo cos θ voy = vo sin θ