# 3.4 Projectile motion  (Page 6/16)

 Page 6 / 16
$R=\frac{{v}_{0}^{2}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}{2\theta }_{0}}{g}\text{,}$

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 [link] .) 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.

## Phet explorations: projectile motion

Blast a Buick out of a cannon! Learn about projectile motion by firing various objects. Set the angle, initial speed, and mass. Add air resistance. Make a game out of this simulation by trying to hit a target.

## Summary

• Projectile motion is the motion of an object through the air that is subject only to the acceleration of gravity.
• To solve projectile motion problems, perform the following steps:
1. 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.
2. Analyze the motion of the projectile in the horizontal direction using the following equations:
$\text{Horizontal motion}\left({a}_{x}=0\right)$
$x={x}_{0}+{v}_{x}t$
${v}_{x}={v}_{0x}={\mathbf{\text{v}}}_{\text{x}}=\text{velocity is a constant.}$
3. Analyze the motion of the projectile in the vertical direction using the following equations:
$\text{Vertical motion}\left(\text{Assuming positive direction is up;}\phantom{\rule{0.25em}{0ex}}{a}_{y}=-g=-9\text{.}\text{80 m}{\text{/s}}^{2}\right)$
$y={y}_{0}+\frac{1}{2}\left({v}_{0y}+{v}_{y}\right)t$
${v}_{y}={v}_{0y}-\text{gt}$
$y={y}_{0}+{v}_{0y}t-\frac{1}{2}{\text{gt}}^{2}$
${v}_{y}^{2}={v}_{0y}^{2}-2g\left(y-{y}_{0}\right).$
4. Recombine the horizontal and vertical components of location and/or velocity using the following equations:
$s=\sqrt{{x}^{2}+{y}^{2}}$
$\theta ={\text{tan}}^{-1}\left(y/x\right)$
$v=\sqrt{{v}_{x}^{2}+{v}_{y}^{2}}$
${\theta }_{\text{v}}={\text{tan}}^{-1}\left({v}_{y}/{v}_{x}\right).$
• The maximum height $h$ of a projectile launched with initial vertical velocity ${v}_{0y}$ is given by
$h=\frac{{v}_{0y}^{2}}{2g}.$
• The maximum horizontal distance traveled by a projectile is called the range . 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
$R=\frac{{v}_{0}^{2}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}{2\theta }_{0}}{g}.$

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