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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?

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Mathematica is a good choice demonstrations.wolfram.com/… They also have packages about this, library.wolfram.com/infocenter/Articles/2698 –  Marvin Jan 26 '13 at 2:00
The air resistance is proportional to $v^2$ for all but the lowest velocities, so your equation isn't a good model for the real world. There is no closed form solution for the general case, though for vertical motion the equation can be solved. See hyperphysics.phy-astr.gsu.edu/hbase/mechanics/quadrag.html –  John Rennie Jan 26 '13 at 10:07

2 Answers 2

up vote 5 down vote accepted

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 $\lambda^* \to \infty$.

enter image description here

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Are you used Buckingham π theorem for making dimensionless coordinates? –  m0nhawk Jan 26 '13 at 9:24
@m0nhawk Not really. I just chose the obvious units for velocity and time, and then $\lambda$ forms on its own. But you could get it that way too. –  Jaime Jan 27 '13 at 9:20

I feel tempted to give a bit of a twist to this question. As John Rennie mentions in a comment above, unless you are talking about slowly moving microscopic projectiles, drag forces are best modeled as being proportional to the square of the velocity.

This brings me to a human-interest story relevant to the problem under discussion.

Last summer, it was reported that a 16 year old Indian schoolboy had solved a 350 year old problem that had stumped researchers since the time of Newton. All of this was highly over-hyped. The story, however, is relevant to the present discussion as one of the two problems solved by the boy is that of the movement of a particle under a uniform gravitational acceleration and a quadratic deceleration due to air resistance. You can find a Physics SE discussion here, and comments from professors at Dresden University (where the schoolboy did an internship) can be found here.

the 16 year old schoolboy, Shouryya Ray, with a poster highlighting a constant of motion for a ballistic particle undergoing a quadratic drag force

As OP is "seeking fun in solving equations", it might be worthwhile to check out some of the above links.

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