13.3: Projectile Problem
- Page ID
- 92197
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Recall the problem from Section 12.6 of directing a projectile to hit a target on a hill at position ( \(\left.x_{t}, y_{t}\right)\), when the muzzle velocity is fixed and we're allowed to vary the angle. We found that in order to hit the target, the launch angle \(\theta\) is the solution to Eq. 12.6.9,
\[x_{t} \sin 2 \theta-2 y_{t} \cos ^{2} \theta=\frac{g x_{t}^{2}}{v_{0}^{2}}\]
which cannot be solved in closed form and must be solved numerically. To solve this for \(\theta\) using Newton's method, we must first put it in the form \(f(\theta)=0\) :
\[f(\theta)=x_{t} \sin 2 \theta-2 y_{t} \cos ^{2} \theta-\frac{g x_{t}^{2}}{v_{0}^{2}}=0\]
Newton's method also requires the derivative of \(f\) :
\[
\begin{aligned}
f^{\prime}(\theta) & =2 x_{t} \cos 2 \theta+4 y_{t} \cos \theta \sin \theta \\[6pt]
& =2 x_{t} \cos 2 \theta+2 y_{t} \sin 2 \theta
\end{aligned}
\]
Using these expressions for \(f(\theta)\) and \(f^{\prime}(\theta)\) in Newton's method (Eq. (13.1.2)), we find an iterative expression that lets us solve numerically for the launch angle \(\theta\) :
\[\theta_{n+1}=\theta_{n}-\frac{x_{t} \sin 2 \theta_{n}-2 y_{t} \cos ^{2} \theta_{n}-g x_{t}^{2} / v_{0}^{2}}{2 x_{t} \cos 2 \theta_{n}+2 y_{t} \sin 2 \theta_{n}}\]
Here the target coordinates \(\left(x_{t}, y_{t}\right)\) are known, as are the muzzle velocity \(v_{0}\) and acceleration due to gravity \(g\), so the only variable on the right-hand side is \(\theta_{n}\). To use this expression, we begin with an initial guess for the launch angle, \(\theta_{0}\) (in radians). Then plug this \(\theta_{0}\) into the right-hand side, which returns \(\theta_{1}\); for the next iteration, plug this \(\theta_{1}\) into the right-hand side, which returns \(\theta_{2}\), etc. After a few iterations, you should get approximately the same angle over and over again on successive iterations. If the target is out of range, the method will "blow up" and not converge, typically by returning larger and larger values of \(\theta_{n}\) for each iteration.
For this type of iterative calculation, it is handy to program the iteration formula into a programmable calculator, or write a computer program.