Projectile with air resistance. Maximum range angle.

The equations of motion with initial conditions are: \begin{align*} \ddot{y} &= -g-\gamma \dot{y}\\ \ddot{x} &= -\gamma \dot{x}\\ \dot{y}(0) &= V \sin(\alpha)\\ \dot{x}(0) &= V \cos(\alpha)\\ y(0)&=x(0)=0~~. \end{align*} We define \begin{equation}\eta=\frac{g}{V\gamma}=\frac{V_\infty}{V}~~,\end{equation} where \(V_\infty=g/\gamma\) is the so-called terminal speed. Rescaling \(t\to (\gamma t)\), \(x\to x\gamma/V\), and \(y\to y\gamma/V\), we are left with unitless equations: \begin{align} \begin{split} \ddot{y} &= -\eta-\dot{y}\\ \ddot{x} &= - \dot{x}\\ \dot{y}(0) &= \sin(\alpha)\\ \dot{x}(0) &= \cos(\alpha)\\ y(0)&=x(0)=0~~. \end{split}\end{align} The solution is given by (see here) \begin{align} y(t) &= \left(1-e^{-t}\right) (\sin (\alpha )+\eta )-\eta t\\ x(t) &= \left(1-e^{-t}\right) \cos (\alpha )~~. \end{align} These effectively give the \((x,y)\) parametric trajectories of the projectiles respect to time.

Determining which is the angle which gives the maximum range of the projectile is a bit algebraically involved (in a similar way as this problem for example). To tackle this problem, we will calculate the envelope of the trajectories while varying \(\alpha\), the throwing angle. In projectile problems, this is sometimes called the "envelope of safety."

Figure 1: Many trajectories and their envelope in pink for \(\eta=0.03\). The intersection of the envelope with the \(x\) axis is the maximum range.

If the position of the projectile is given by \(\vec{r}(t,\alpha)=(x(t,\alpha),y(t,\alpha))\) the envelope is given by the condition \[ \dot{\vec{r}} \parallel \partial_\alpha\vec{r}\] which translates to \begin{equation} e^{-2 t} \left(e^t-1\right) \left(\eta \left(e^t-1\right) \sin (\alpha )-1\right)=0~~, \end{equation} which clearly implies \begin{equation}t(\alpha)=\log \left(\frac{\csc (\alpha )}{\eta }+1\right)~~.\end{equation} Thus, the envelope is given parametrically by \begin{equation} \vec{r}\left(t(\alpha),\alpha\right)=\left(\frac{\cot (\alpha )}{\csc (\alpha )+\eta },\frac{1+\eta\csc(\alpha) }{\eta +\csc(\alpha)}-\eta \log \left(\frac{\csc (\alpha )+\eta }{\eta }\right)\right)~~. \end{equation} To find the maximum range, we need the conditions associated with \[y=0\Longleftrightarrow \frac{1+\eta\csc(\alpha) }{\eta +\csc(\alpha)} = \eta \log \left(\frac{\csc (\alpha )+\eta }{\eta }\right)\].

To advance further we will define \(\epsilon=\eta+\csc(\alpha)\). Then, the equation above is re-written as: \begin{equation} \begin{split} \eta \log\left(\frac{\epsilon}{\eta}\right) &= \frac{\eta}{\epsilon}(\epsilon-\eta)+\frac{1}{\epsilon}\\ \log\left(\frac{\epsilon}{e\eta}\right) &= \frac{1-\eta^2}{\eta\epsilon} \\ \to \frac{\epsilon}{e\eta} &= e^{\frac{1-\eta^2}{\eta\epsilon}}\\ \frac{\epsilon\eta}{(1-\eta^2)}\frac{(1-\eta^2)}{e\eta^2} &= e^{\frac{1-\eta^2}{\eta\epsilon}}\\ \frac{(1-\eta^2)}{e\eta^2} &= \frac{(1-\eta^2)}{\eta\epsilon} e^{\frac{1-\eta^2}{\eta\epsilon}}\\ \to \frac{(1-\eta^2)}{\eta\epsilon} &= W\left(\frac{(1-\eta^2)}{e\eta^2}\right)\\ \epsilon = \csc(\alpha)+\eta &= \frac{(1-\eta^2)}{\eta W\left(\frac{(1-\eta^2)}{e\eta^2}\right)}\\ \longrightarrow \quad \boxed{\csc(\alpha) = \frac{(1-\eta^2)}{\eta W\left(\frac{(1-\eta^2)}{e\eta^2}\right)} -\eta} \end{split}~~, \end{equation} where \(W\) is the Lambert function. This is a well studied function, although not an elementary one. Note that since \(\eta = \frac{g}{V\gamma}\), we have that the no-resistance limit of the equation is obtained by taking \(\eta\to+\infty\). The argument of the Lambert function goes to \(-e^{-1}\), where the Lambert function has a singular point but is continuous and its value is -1. Asymptotic expansions of \(W\) around this point are derived in arXiv1209.0735 and are given in this case by \begin{equation}\csc(\alpha) = \sqrt{2}+ \frac{1}{3 \eta }+O(\eta^{-2})\approx \sqrt{2}+\frac{V\gamma}{3g}=\sqrt{2}+\frac{V}{3V_\infty}~~.\end{equation} We see that we recover the no resistance well-known result (\(\csc(45^\circ)=\sqrt{2}\)). In general, the angle of maximum range in the presence of air damping linear in velocity is smaller than 45\(^\circ\). The limit of no air resistance is attained in the low velocity regimes as well.

Projectile with air resistance. Maximum range angle.

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