11.1 Introduction to Projectile Motion
Projectile motion is a fundamental concept in physics that deals with the motion of an object under the influence of gravity while experiencing no other forces once set in motion. This type of motion is very familiar to us all as we often observe it in everyday life. Shooting a basketball and throwing a TV remote on the couch are just two examples. Though totally different actions, the math describing the physical motion of both of these is exactly the same. We often use the word “trajectory” to describe the path of an object experiencing projectile motion.
In this section, we will break down the key elements of projectile motion, the equations that govern it, and how to solve problems related to this concept.
10.2 Horizontal and Vertical Motion
Projectile motion can be broken down into two independent components: horizontal and vertical motion. The horizontal motion is characterized by a constant velocity since there are no forces acting on the object in this direction. In contrast, the vertical motion is influenced by gravity, which causes the object to accelerate downward.
| Horizontal Motion: | Constant Velocity |
| Vertical Motion: | Constant Acceleration |
To analyze projectile motion, we must consider these two components separately. Generally, we start by solving for the vertical motion first. Then, we move on to the horizontal motion. However, this is not always the case, so it is important to read the question carefully.
10.2.1 Horizontal motion:
Displacement (
): 
Velocity (
): 
(constant)
Acceleration (
): 
10.2.2 Vertical motion:
Displacement (
): 
Velocity (
): 
Acceleration (
): 
or 
(where g is the acceleration due to gravity, approximately 9.81 m/s^2)
A note on signs: You decide which direction is positive and which is negative. As long as you are consistent with your choice as you go about solving a problem, the math will work out. Suppose you choose downward to be the “positive” direction. In this case, an object experiencing projectile motion will have a positive acceleration in the y-direction because it is moving faster and faster in the “positive” direction. (This is a popular choice of coordinate system for this type of problem.) On the other hand, suppose you choose downward to be the “negative” direction. To account for this, your acceleration must be negative as the object is moving with increasing speed in the “negative” direction.
10.3 Projectile Motion Problem Solving
Let us consider the most common type of projectile motion problem.
Example:
You launch a ball at an angle of 30 degrees with respect to the horizontal axis. The initial speed is 12 meters per second. You are standing on top of a building that is 23 meters above the ground. How long does it take for the ball to reach the ground on the street below? How far away from the foot of the building does the ball land?
Step 1: Break up the initial velocity into x and y components. Use trigonometry to solve.
NOTE: What if we launched the ball at an angle of 30 degrees below the horizontal? How would that affect the answers above? In that case, would remain unchanged, but 
would become negative (the triangle above would flip upside down). We chose down to be the negative direction when we set up this problem. So, a ball thrown slightly downward must have a negative initial velocity in the y-direction.
Step 2: Solve for the y-direction motion. Specifically, we want to know how long the object is in the air. Y-direction motion is constant acceleration motion, so we will use those equations to solve for t.
The 
in the first line above tells us that the equation is a quadratic equation (the highest power in the exponent is “2”). These types of equations are most easily solved graphically. To do so, move all of the terms over to the same side of the equals sign (line 3 above).
Graph the equation (i.e., 
). Once you obtain this graph, find where it crosses the horizontal axis (this is where 
). It will do this in two different locations (shown below). The positive value is the one we want because a negative time value doesn’t make any sense in this situation.
As you can see, the time it takes for the object to hit the ground is 2.864 seconds.
A word of caution: It may be tempting to think that the graph above represents the trajectory of the ball. It does not. This is a graph of vertical height vs. time. We would need to graph horizontal height as a function of distance traveled (y vs. x) in order to show the trajectory. So, keep in mind that the horizontal units on this graph are “seconds,” not “meters.” It’s a common mistake to think that this graph shows a horizontal distance traveled of 2.864 meters, but this is not correct. It shows a time of flight of 2.864 seconds.
Step 3: Solve for the x-direction motion. Now we wish to determine how far the ball traveled. X-direction motion is constant velocity motion. So, we can quickly calculate the distance traveled by using the value of from step 1 and the value of 
from step 2 as shown below.
Remember that you are working in units of meters! Your final answer thus turns out to be:
seconds, and
meters.
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Continue to Chapter 12: Uniform Circular Motion
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