12.1: Introduction to Projectile Motion
- Page ID
- 99781
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)An important example of two-dimensional motion under constant acceleration is the motion of a projectile (e.g. a cannonball fired from a cannon) at the surface of the Earth (Fig. \(\PageIndex{1}\)).
The acceleration in this case is the acceleration due to gravity, so the constant-acceleration equations apply. The position vector as a function of time is given by Eq. (11.2.7):
\[\mathbf{r}(t)=\frac{1}{2} \mathbf{a} t^{2}+\mathbf{v}_{0} t+\mathbf{r}_{0}\]
where \(\mathbf{v}_{0}\) is the initial velocity of the cannonball, called the muzzle velocity. Let's take time \(t=0\) to be the instant the cannonball leaves the cannon. Then if we choose the origin to be at the cannon (Fig.\(\PageIndex{1}\)), then \(\mathbf{r}_{0}=\mathbf{0}\). The acceleration in this case is in the \(-y\) direction, so \(\mathbf{a}=-g \mathbf{j}\), and Eq. \(\PageIndex{1}\) becomes
\[\mathbf{r}(t)=-\frac{1}{2} g t^{2} \mathbf{j}+\mathbf{v}_{0} t\]
where the initial velocity \(\mathbf{v}_{0}=v_{x 0} \mathbf{i}+v_{y 0} \mathbf{j}\). This vector equation actually represents two scalar equations: one for \(x(t)\) and one for \(y(t)\) :
\[
\begin{align}
& x(t)=v_{x 0} t \\[6pt]
& y(t)=-\frac{1}{2} g t^{2}+v_{y 0} t
\end{align}
\]
Typically in real life you will not know the cartesian components of the velocity vector ( \(v_{x 0}\) and \(v_{y 0}\) ); instead you are more likely to know the magnitude of the muzzle velocity \(v_{0}\) and the launch angle \(\theta\). Converting the muzzle velocity vector from rectangular to polar form,
\[
\begin{align}
& v_{0 x}=v_{0} \cos \theta \\[6pt]
& v_{0 y}=v_{0} \sin \theta
\end{align}
\]
Equations \(\PageIndex{3}\) and \(\PageIndex{4}\) then become
\[
\begin{align}
& x(t)=\left(v_{0} \cos \theta\right) t \\[6pt]
& y(t)=-\frac{1}{2} g t^{2}+\left(v_{0} \sin \theta\right) t
\end{align}
\]
These equations give the \(x\) and \(y\) coordinates of the projectile at any time \(t\).
Now let's consider a few questions we can ask about the motion of a projectile.