If you throw a projectile from the ground at a certain angle, it's not hard to see that, assuming we're in a vacuum, throwing it at 45º from the ground will always make it go farthest before it hits the ground again. I wanted to find out what's the optimal angle if we take into account air resistance.
Before I go on, let me say that there's really no point to this question. I'm not trying to solve a practical problem. I just thought it would be fun to try to solve the equations.
This is the model I'll use: We'll assume that air resistance is roughly proportional to velocity (aka. Stokes drag; this fails for high speeds/Reynolds numbers, so let's not go there); in symbols, $$\mathbf{F} = -k\mathbf{v}.$$ We'll throw the projectile with mass $m$ from $(0,0)$ with initial speed $v_0$, forming an angle $\alpha$ with the ground, and let's say $\gamma = k/m$ because it will turn up frequently.
The differential equations are:
$$\begin{align} \ddot{x} + \gamma \dot{x} &= 0 \\ \ddot{y} + \gamma \dot{y} &= -g \end{align}$$
with initial conditions $(x(0), y(0)) = (0,0)$ and $(\dot{x}(0), \dot{y}(0) )= v_0(\cos \alpha, \sin \alpha)$. Solving (assuming I haven't made any mistakes), we get:
$$ \begin{align} x &= \frac{v_0}{\gamma}\cos \alpha (1- e^{-\gamma t}) \\ y &= (\frac{v_0}{\gamma} \sin \alpha + \frac{g}{\gamma^2})(1-e^{-\gamma t}) - \frac{g}{\gamma}t. \end{align} $$
Now the thing to do would be to solve $y(t) = 0$, substitute that into $x$, differentiate with respect to $\alpha$ and set it equal to $0$. However, that gets messy quickly, because the solution involves the Lambert-W function and it's all a mess. I haven't even tried to substitute into $x(t)$.
So finally, my question is: is there a simpler or numerical way to solve this? Is there even a single angle which will always work, or does it depend on the conditions? Can we find that out without actually solving?
Answer
This is one of those problems that take on a much neater solution if you go dimensionless...
If you take a new time, $\tau$, horizontal coordinate, $\xi$, and vertical coordinate, $\eta$, such that
$$\tau = t \gamma,\ \xi = x \frac{\gamma}{v_0 \cos \alpha},\ \eta = y \frac{\gamma}{v_0 \sin \alpha}$$
you can rewrite your equations as
$$\ddot{\xi} + \dot{\xi} = 0$$ $$\ddot{\eta} + \dot{\eta} = -\frac{g}{\gamma v_0 \sin\alpha} = -\lambda$$
with initial conditions at $t=0$
$$\xi = \eta = 0,\ \dot{\xi} = \dot{\eta} = 1$$
This has solutions
$$\xi = 1 -e^{-\tau},\ \eta = (1+\lambda)(1 -e^{-\tau}) - \lambda \tau$$
and setting $\eta = 0$, we can get rid of $\tau$, and the range will be $\xi^*$ that fulfills
$$-\frac{1+\lambda}{\lambda}\xi^* = \log (1-\xi^*)$$
The maximum range will happen when $d\xi^*/d\alpha = 0$, so it is easy to show that at the maximum range:
$$\frac{d}{d\alpha}\frac{1}{\lambda} = \frac{1}{\lambda \tan \alpha}$$ $$\frac{d\xi^*}{d\alpha} = \xi^* \tan \alpha$$
Differentiating the equation for $\xi^*$ w.r.t. $\alpha$ using these two last formulas will, after a lot of cancelling and rearranging, take you to
$$\xi^* = \frac{1}{1+\lambda \sin^2 \alpha}$$
You can use this relation in the first expression we got for $\xi^*$ to get rid of $\xi^*$ and get a trascendental equation for $\alpha$ that looks something like:
$$\frac{\sin \alpha + \lambda^*}{\lambda^*}\frac{1/\lambda^*}{1/\lambda^* + \sin \alpha} = \log(1/\lambda^* +\sin\alpha) - \log(\sin \alpha)$$
where $\lambda^* = \lambda \sin \alpha$ and is independent of $\alpha$. This last equation you would want to solve numerically to get $\alpha^*$, the angle producing the maximum range.
While messy, the solution clearly depends on $\lambda^*$, so there will not be a single angle that always works.
EDIT Just run this last equation through a numerical simulation, and here's the optimal angle $\alpha^*$ (actually $\sin\alpha^*$) as a function of $\lambda^*$, showing how it grows ever closer to $\sqrt{2}/2$ for $\lambda^* \to \infty$.
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