<h2>Summary</h2>
<ul>
<li>Projectile motion is the motion of an object through the air that is subject only to the acceleration of gravity.</li>
<li>To solve projectile motion problems, perform the following steps:

<ol type="1">
<li>Determine a coordinate system. Then, resolve the position and/or velocity of the object in the horizontal and vertical components. The components of position \(\mathbf{s}\) are given by the quantities \(x\) and \(y\), and the components of the velocity \(\mathbf{v}\) are given by \({v}_{x}=v\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta \) and \({v}_{y}=v\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta \), where \(v\) is the magnitude of the velocity and \(\theta \) is its direction.</li>
<li>Analyze the motion of the projectile in the horizontal direction using the following equations:

<div>\(\text{Horizontal motion}\left({a}_{x}=0\right)\)</div>
<div>\(x={x}_{0}+{v}_{x}t\)</div>
<div>\({v}_{x}={v}_{0x}={\mathbf{\text{v}}}_{\text{x}}=\text{velocity is a constant.}\)</div>
</li>
<li>Analyze the motion of the projectile in the vertical direction using the following equations:

<div>\(\text{Vertical motion}(\text{Assuming positive direction is up;}\phantom{\rule{0.25em}{0ex}}{a}_{y}=-g=-9\text{.}\text{80 m}{\text{/s}}^{2})\)</div>
<div>\(y={y}_{0}+\cfrac{1}{2}\left({v}_{0y}+{v}_{y}\right)t\)</div>
<div>\({v}_{y}={v}_{0y}-\text{gt}\)</div>
<div>\(y={y}_{0}+{v}_{0y}t-\cfrac{1}{2}{\text{gt}}^{2}\)</div>
<div>\({v}_{y}^{2}={v}_{0y}^{2}-2g\left(y-{y}_{0}\right).\)</div>
</li>
<li>Recombine the horizontal and vertical components of location and/or velocity using the following equations:

<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>\({\theta }_{\text{v}}={\text{tan}}^{-1}\left({v}_{y}/{v}_{x}\right).\)</div>
</li>
</ol>
</li>
<li>The maximum height \(h\) of a projectile launched with initial vertical velocity \({v}_{0y}\) is given by

<div>\(h=\cfrac{{v}_{0y}^{2}}{2g}.\)</div>
</li>
<li>The maximum horizontal distance traveled by a projectile is called the <strong>range</strong>. The range \(R\) of a projectile on level ground launched at an angle \({\theta }_{0}\) above the horizontal with initial speed \({v}_{0}\) is given by

<div>\(R=\cfrac{{v}_{0}^{2}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}{2\theta }_{0}}{g}.\)</div>
</li>
</ul>
<h2>Glossary</h2>
<h3>air resistance</h3>
<p>a frictional force that slows the motion of objects as they travel through the air; when solving basic physics problems, air resistance is assumed to be zero</p>
<h3>kinematics</h3>
<p>the study of motion without regard to mass or force</p>
<h3>motion</h3>
<p>displacement of an object as a function of time</p>
<h3>projectile</h3>
<p>an object that travels through the air and experiences only acceleration due to gravity</p>
<h3>projectile motion</h3>
<p>the motion of an object that is subject only to the acceleration of gravity</p>
<h3>range</h3>
<p>the maximum horizontal distance that a projectile travels</p>
<h3>trajectory</h3>
<p>the path of a projectile through the air</p>
<section data-element-type="conceptual-questions"></section>

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Summarizing 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.

Physics

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.


Summarizing Projectile Motion

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