12.5: Hitting a Target on the Ground
- Page ID
- 92188
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Now let's look at the problem of using a projectile to hit a target on the ground at range \(R\). We could do this by fixing the muzzle velocity and varying the launch angle, or by fixing the launch angle and varying the muzzle velocity, or by varying both.
Fixed Launch Angle
The less common situation is to fix the launch angle \(\theta\) and allow the muzzle velocity \(v_{0}\) to vary. Beginning with Eq. (12.2.6),
\[R=\frac{v_{0}^{2}}{g} \sin 2 \theta\]
we solve for muzzle velocity:
\[v_{0}=\sqrt{\frac{g R}{\sin 2 \theta}}\]
There will always be a solution to this equation unless \(\theta \geq 90^{\circ}\), which corresponds to pointing the cannon backwards. In this case \(v_{0}\) will be imaginary, and there is no muzzle velocity that will allow the projectile to reach the target.
Suppose we have a cannon fixed at an angle of \(25^{\circ}\) and wish to hit a target at a distance of \(R=250 \mathrm{~m}\). What muzzle velocity \(v_{0}\) is required?
Solution
By Eq. \(\PageIndex{2}\),
\[v_{0}=\sqrt{\frac{g R}{\sin 2 \theta}}=56.55 \mathrm{~m} / \mathrm{s} \]
Fixed Muzzle Velocity
The more common situation is trying to hit a target when the muzzle velocity is fixed and the launch angle is allowed to vary. In this case we solve Eq. (9.1.6) for \(\theta\) :
\[\theta=\frac{1}{2} \sin ^{-1}\left(\frac{g R}{v_{0}^{2}}\right)\]
Suppose the muzzle velocity is \(v_{0}=40 \mathrm{~m} / \mathrm{s}\) and the target is at a distance of \(R=75 \mathrm{~m}\). What launch angle is needed to hit the target?
Solution
The launch angle is given by
\[\theta=\frac{1}{2} \sin ^{-1}\left(\frac{g R}{v_{0}^{2}}\right)=13.67^{\circ} \text { and } 76.33^{\circ}\]
Recall that the arcsine of a number returns two angles in the range \([0,2 \pi)\), so there will generally be two solutions to Eq. \(\PageIndex{4}\). In this example, the "calculator" solution is \(13.67^{\circ}\), and other solution is \(76.33^{\circ}\). In general, there will be two complementary launch angles that will both hit the target.
As a second example, suppose the muzzle velocity is \(v_{0}=40 \mathrm{~m} / \mathrm{s}\) and the target is at a distance of \(R=200 \mathrm{~m}\). What launch angle is needed to hit the target?
Solution
The launch angle is given by
\[\theta=\frac{1}{2} \sin ^{-1}\left(\frac{g R}{v_{0}^{2}}\right)=\frac{1}{2} \sin ^{-1} 1.225=? ? ?\]
The arcsine of a number greater than 1 is not defined \({ }^{2}\), so \(\theta\) cannot be found in this case. Physically, this means that the target is out of range at this muzzle velocity. For a muzzle velocity of \(40 \mathrm{~m} / \mathrm{s}\), the maximum range is for \(\theta=45^{\circ}\), which by Eq. (12.2.6) is \(163 \mathrm{~m}\) - so \(200 \mathrm{~m}\) is out of range.