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<p><strong>Projectile motion</strong> is the <strong>motion</strong> of an object thrown or projected into the air, subject to only the acceleration of gravity. The object is called a <strong>projectile</strong>, and its path is called its <strong>trajectory</strong>.</p>
<p>The motion of falling objects, as covered in the <a data-page="13" href="https://app.ulearngo.com/tutorial/physics-201/">previous tutorial</a>, 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 <strong>air resistance </strong><em>is negligible</em>.</p>
<p>The most important fact to remember here is that <em>motions along perpendicular axes are independent</em> 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.</p>
<p>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 <em>x</em>-axis and the vertical axis the <em>y</em>-axis.</p>
<p>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.</p>
<p>The magnitudes of these vectors are <em>s</em>, <em>x</em>, and <em>y</em>. (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 \(\mathbf{y}\).)</p>
<p>Of course, to describe motion we must deal with velocity and acceleration, as well as with displacement. We must find their components along the <em>x</em>- and <em>y</em>-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: \({a}_{y}=–g=–9.80 m{\text{/s}}^{2}\).</p>
<p>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.</p>
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<h3 data-type="title">Review of Kinematic Equations (constant \(a\))</h3>
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<p>\(x={x}_{0}+\stackrel{-}{v}t\)</p>
<p>\(\stackrel{-}{v}=\frac{{v}_{0}+v}{2}\)</p>
<p>\(v={v}_{0}+\text{at}\)</p>
<p>\(x={x}_{0}+{v}_{0}t+\frac{1}{2}{\text{at}}^{2}\)</p>
<p>\({v}^{2}={v}_{0}^{2}+2a\left(x-{x}_{0}\right)\text{.}\)</p>
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<div class="wp-caption alignnone" style="width: 800px"><img alt="A soccer player is kicking a soccer ball. The ball travels in a projectile motion and reaches a point whose vertical distance is y and horizontal distance is x. The displacement between the kicking point and the final point is s. The angle made by this displacement vector with x axis is theta." height="460" src="https://data.ulearngo.com/assets/bd0f84b3-eb28-4852-beca-e4728f9d27a1" width="800"/><p class="wp-caption-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.</p></div>
<p>Given these assumptions, the following steps are then used to analyze projectile motion:</p>
<h3><em><strong>Step 1. </strong></em><em>Resolve or break the motion into horizontal and vertical components along the x- and y-axes.</em></h3>
<p>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 \(\mathrm{y.}\)</p>
<p>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.</p>
<h3><em><strong>Step 2. </strong></em><em>Treat the motion as two independent one-dimensional motions, one horizontal and the other vertical.</em></h3>
<p>The kinematic equations for horizontal and vertical motion take the following forms:</p>
<div>\(\text{Horizontal Motion}\left({a}_{x}=0\right)\)</div>
<div>\(x={x}_{0}+{v}_{x}t\)</div>
<div>\({v}_{x}={v}_{0x}={v}_{x}=\text{velocity is a constant.}\)</div>
<div>\(\text{Vertical Motion}\left(\text{assuming positive is up}\phantom{\rule{0.25em}{0ex}}{a}_{y}=-g=-9.\text{80}{\text{m/s}}^{2}\right)\)</div>
<div>\(y={y}_{0}+\frac{1}{2}\left({v}_{0y}+{v}_{y}\right)t\)</div>
<div>\({v}_{y}={v}_{0y}-\text{gt}\)</div>
<div>\(y={y}_{0}+{v}_{0y}t-\frac{1}{2}{\mathrm{gt}}^{2}\)</div>
<div>\({v}_{y}^{2}={v}_{0y}^{2}-2g\left(y-{y}_{0}\right)\text{.}\)</div>
<h3><em><strong>Step 3.</strong></em><em> Solve for the unknowns in the two separate motions—one horizontal and one vertical.</em></h3>
<p>Note that the only common variable between the motions is time \(t\). The problem solving procedures here are the same as for one-dimensional <strong>kinematics</strong> and are illustrated in the solved examples below.</p>
<h3><em><strong>Step 4. </strong></em><em>Recombine the two motions to find the total displacement </em>\(\mathbf{\text{s}}\)<em> and velocity </em>\(\mathbf{\text{v}}\).</h3>
<p>Because the <em>x</em> - and <em>y</em> -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 \(\mathbf{v}\):</p>
<p><strong>Total displacement and velocity</strong></p>
<div>\(s=\sqrt{{x}^{2}+{y}^{2}}\)</div>
<div>\(\theta ={\text{tan}}^{-1}\left(y/x\right)\)</div>
<div>\(v=\sqrt{{v}_{x}^{2}+{v}_{y}^{2}}\)</div>
<div>\({\theta }_{v}={\text{tan}}^{-1}\left({v}_{y}/{v}_{x}\right)\text{.}\)</div>
<div class="wp-caption alignnone" style="width: 800px"><img alt="In part a the figure shows projectile motion of a ball with initial velocity of v zero at an angle of theta zero with the horizontal x axis. The horizontal component v x and the vertical component v y at various positions of ball in the projectile path is shown. In part b only the horizontal velocity component v sub x is shown whose magnitude is constant at various positions in the path. In part c only vertical velocity component v sub y is shown. The vertical velocity component v sub y is upwards till it reaches the maximum point and then its direction changes to downwards. In part d resultant v of horizontal velocity component v sub x and downward vertical velocity component v sub y is found which makes an angle theta with the horizontal x axis. The direction of resultant velocity v is towards south east." height="1181" src="https://data.ulearngo.com/assets/7a0b826c-5e8f-4f2f-acab-40d9b2beb515" width="800"/><p class="wp-caption-text">(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.</p></div>

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Projectile Motion

Physics is a science discipline concerned with the nature and properties of matter and energy. Physics topics include mechanics, heat, light and radiation, sound, electricity, magnetism, the structure of atoms, and so on.

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Explore the principles of two-dimensional kinematics, including mastering vector addition and subtraction, resolving a vector into components, and analyzing projectile motion and relative velocities.


Projectile Motion

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