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.
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 the previous tutorial, 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 at the start of this tutorial, 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.
The figure below 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: \({a}_{y}=–g=–9.80 m{\text{/s}}^{2}\).
Note that this definition above 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.
Review of Kinematic Equations (constant \(a\))
\(x={x}_{0}+\stackrel{-}{v}t\)
\(\stackrel{-}{v}=\frac{{v}_{0}+v}{2}\)
\(v={v}_{0}+\text{at}\)
\(x={x}_{0}+{v}_{0}t+\frac{1}{2}{\text{at}}^{2}\)
\({v}^{2}={v}_{0}^{2}+2a\left(x-{x}_{0}\right)\text{.}\)
The total displacement \(\mathbf{s}\) of a soccer ball at a point along its path. The vector \(\mathbf{s}\) has components \(\mathbf{x}\) and \(\mathbf{y}\) along the horizontal and vertical axes. Its magnitude is \(s\), and it makes an angle \(\theta \) with the horizontal.
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 \({A}_{x}=A\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta \) and \({A}_{y}=A\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta \) 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 \({v}_{x}=v\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta \) and \({v}_{y}=v\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\mathrm{\theta ,}\) where \(v\) is the magnitude of the velocity and \(\theta \) is its direction, as shown in the figure below. 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 previous lessons on analytical methods of vector addition and subtraction and employing \(A=\sqrt{{A}_{x}^{2}+{A}_{y}^{2}}\) and \(\theta ={\text{tan}}^{-1}\left({A}_{y}/{A}_{x}\right)\) 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
(a) We analyze two-dimensional projectile motion by breaking it into two independent one-dimensional motions along the vertical and horizontal axes. (b) The horizontal motion is simple, because \({a}_{x}=0\) and \({v}_{x}\) is thus constant. (c) The velocity in the vertical direction begins to decrease as the object rises; at its highest point, the vertical velocity is zero. As the object falls towards the Earth again, the vertical velocity increases again in magnitude but points in the opposite direction to the initial vertical velocity. (d) The x - and y -motions are recombined to give the total velocity at any given point on the trajectory.
This lesson is part of:
Two-Dimensional Kinematics