3.4 Projectile Motion

Projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. The object is called a projectile, and its path is called its trajectory. The motion of falling objects, as covered in Problem-Solving Basics for One-Dimensional Kinematics, is a simple one-dimensional type of projectile motion in which there is no horizontal movement. In this section, we consider two-dimensional projectile motion, such as that of a football or other object for which air resistance is negligible.

The most important fact to remember here is that motions along perpendicular axes are independent and thus can be analyzed separately. This fact was discussed in Kinematics in Two Dimensions: An Introduction, where vertical and horizontal motions were seen to be independent. The key to analyzing two-dimensional projectile motion is to break it into two motions, one along the horizontal axis and the other along the vertical. (This choice of axes is the most sensible, because acceleration due to gravity is vertical—thus, there will be no acceleration along the horizontal axis when air resistance is negligible.) As is customary, we call the horizontal axis the x-axis and the vertical axis the y-axis. Figure 3.36 illustrates the notation for displacement, where $\mathbf{s}$ is defined to be the total displacement and $\mathbf{x}$ and $\mathbf{y}$ are its components along the horizontal and vertical axes, respectively. The magnitudes of these vectors are s, x, and y. (Note that in the last section we used the notation $\mathbf{A}$ to represent a vector with components ${\mathbf{A}}_x$ and ${\mathbf{A}}_y$. If we continued this format, we would call displacement $\mathbf{s}$ with components ${\mathbf{s}}_x$ and ${\mathbf{s}}_y$. However, to simplify the notation, we will simply represent the component vectors as $\mathbf{x}$ and $\mathbf{y}$.)

Of course, to describe motion we must deal with velocity and acceleration, as well as with displacement. We must find their components along the x- and y-axes, too. We will assume all forces except gravity (such as air resistance and friction, for example) are negligible. The components of acceleration are then very simple: $a_y=–g=–\mathrm{9.80\; m}{\text{/s}}^2$. (Note that this definition assumes that the upwards direction is defined as the positive direction. If you arrange the coordinate system instead such that the downwards direction is positive, then acceleration due to gravity takes a positive value.) Because gravity is vertical, $a_x=0$. Both accelerations are constant, so the kinematic equations can be used.

Given these assumptions, the following steps are then used to analyze projectile motion:

Step 1. Resolve or break the motion into horizontal and vertical components along the x- and y-axes. These axes are perpendicular, so $A_x=A\text{cos}\theta$ and $A_y=A\text{sin}\theta$ are used. The magnitude of the components of displacement $\mathbf{s}$ along these axes are $x$ and $\mathrm{y.}$ The magnitudes of the components of the velocity $\mathbf{v}$ are $v_x=v\text{cos}\theta$ and $v_y=v\text{sin}\mathrm{\theta ,}$ where $v$ is the magnitude of the velocity and $\theta$ is its direction, as shown in Figure 3.37. Initial values are denoted with a subscript 0, as usual.

Step 2. Treat the motion as two independent one-dimensional motions, one horizontal and the other vertical. The kinematic equations for horizontal and vertical motion take the following forms:

Step 3. Solve for the unknowns in the two separate motions—one horizontal and one vertical. Note that the only common variable between the motions is time $t$. The problem solving procedures here are the same as for one-dimensional kinematics and are illustrated in the solved examples below.

Step 4. Recombine the two motions to find the total displacement $\mathbf{\text{s}}$ and velocity $\mathbf{\text{v}}$. Because the x - and y -motions are perpendicular, we determine these vectors by using the techniques outlined in the Vector Addition and Subtraction: Analytical Methods and employing $A=\sqrt{A_x^2+A_y^2}$ and $\theta ={\text{tan}}^{-1}(A_y/A_x)$ in the following form, where $\theta$ is the direction of the displacement $\mathbf{s}$ and ${\theta }_v$ is the direction of the velocity $\mathbf{v}$:

Total displacement and velocity

In solving part (a) of the preceding example, the expression we found for $y$ is valid for any projectile motion where air resistance is negligible. Call the maximum height $y=h$; then,

This equation defines the maximum height of a projectile and depends only on the vertical component of the initial velocity.

One of the most important things illustrated by projectile motion is that vertical and horizontal motions are independent of each other. Galileo was the first person to fully comprehend this characteristic. He used it to predict the range of a projectile. On level ground, we define range to be the horizontal distance $R$ traveled by a projectile. Galileo and many others were interested in the range of projectiles primarily for military purposes—such as aiming cannons. However, investigating the range of projectiles can shed light on other interesting phenomena, such as the orbits of satellites around the Earth. Let us consider projectile range further.

How does the initial velocity of a projectile affect its range? Obviously, the greater the initial speed $v_0$, the greater the range, as shown in Figure 3.40(a). The initial angle ${\theta }_0$ also has a dramatic effect on the range, as illustrated in Figure 3.40(b). For a fixed initial speed, such as might be produced by a cannon, the maximum range is obtained with ${\theta }_0=\text{45º}$. This is true only for conditions neglecting air resistance. If air resistance is considered, the maximum angle is approximately $\text{38º}$. Interestingly, for every initial angle except $\text{45º}$, there are two angles that give the same range—the sum of those angles is $\text{90º}$. The range also depends on the value of the acceleration of gravity $g$. The lunar astronaut Alan Shepherd was able to drive a golf ball a great distance on the Moon because gravity is weaker there. The range $R$ of a projectile on level ground for which air resistance is negligible is given by

where $v_0$ is the initial speed and ${\theta }_0$ is the initial angle relative to the horizontal. The proof of this equation is left as an end-of-chapter problem (hints are given), but it does fit the major features of projectile range as described.

When we speak of the range of a projectile on level ground, we assume that $R$ is very small compared with the circumference of the Earth. If, however, the range is large, the Earth curves away below the projectile and acceleration of gravity changes direction along the path. The range is larger than predicted by the range equation given above because the projectile has farther to fall than it would on level ground. (See Figure 3.41.) If the initial speed is great enough, the projectile goes into orbit. This possibility was recognized centuries before it could be accomplished. When an object is in orbit, the Earth curves away from underneath the object at the same rate as it falls. The object thus falls continuously but never hits the surface. These and other aspects of orbital motion, such as the rotation of the Earth, will be covered analytically and in greater depth later in this text.

Once again we see that thinking about one topic, such as the range of a projectile, can lead us to others, such as the Earth orbits. In Addition of Velocities, we will examine the addition of velocities, which is another important aspect of two-dimensional kinematics and will also yield insights beyond the immediate topic.