TITLE: Angle Maximizing the Distance of a Projectile
QUESTION [22 upvotes]: It is well-known that to maximize the horizontal distance traveled by a projectile fired from the ground at a given speed, one should fire it at a $45^\circ$ angle. What's less-known, though not too difficult to prove with some manipulation of derivatives, is that if one is on an incline of angle $\phi$ below the horizontal, then to maximize horizontal distance traveled, one should fire the projectile at an angle of $\frac{\pi}{4}-\frac{\phi}{2}$ above the horizontal.
To see this, suppose we fire it at an angle of $\theta$ at speed $v_0$ with gravitational constant $g$. Then the equation of the path is $(v_0 t \cos{\theta}, -gt^2+v_0 t \sin{\theta})$, so we are looking for the $t$ at which $v_0 t \cos{\theta} \sin{\phi}-gt^2 \cos{\phi}+v_0 t \sin{\theta} \cos{\phi} = 0$, or $t = \frac{v_0(\cos{\theta}\sin{\phi}+\sin{\theta}\cos{\phi})}{g} = \frac{v_0}{g} \sin(\theta+\phi)$. Then we wish to maximize $\cos{\theta} t$ at this point, or equivalently $\cos{\theta} \sin{(\theta+\phi)}$. Taking the derivative with respect to $\theta$, we find $-\sin{\theta} \sin{\theta+\phi}+\cos{\theta} \cos(\theta+\phi) = \cos{(2\theta+\phi)} = 0$. Thus $2\theta+\phi = \frac{\pi}{2}$, giving us our answer.
Since this formula looks so nice, I'm wondering whether there is a nicer proof of this fact, possibly using symmetry and/or with more physical intuition. In particular, it might involve some kind of rotation (maybe by $\frac{\phi}{2}$?).

REPLY [5 votes]: I've given two answers here earlier, but thought I'd add a third.  This one focuses on the OP's original request for an argument that utilizes symmetry.  It also avoids doing any explicit algebraic calculations for projectile motion, using instead just general principles and physical intuition.
The key lies in the fact that the answer to the problem is the angle bisector between the vertical (i.e., the direction of gravity) and the slope of the ground.  This suggests that one should be able to use the angle bisector as an axis of mirror-image symmetry.  But there's an obvious problem:  Gravity and ground are not physically symmetric.  The angle bisector is not an axis of symmetry for the parabola of projectile motion.
The solution to this is equally obvious (at least in hindsight):  Pretend that gravity and ground are symmetric. That is, go ahead and take the asymmetric picture, symmetrize it with a reflection across the angle bisector, and then try to attach some meaning to things.
It's helpful to start with the classic flat-ground case.
Draw any (downward-pointing) parabola that leaves from the origin (with positive slope) and passes through the $x$-axis again at some point, say $(D,0)$.  Go ahead and extend the drawing past $D$ (we'll see why in a couple of paragraphs).  Note that physics (or the geometry) tells you that the slope of the parabola at $D$ is the negative of its slope at the origin.
Now reflect this drawing through the line $y=x$ (i.e., the angle bisector between the horizontal and the vertical), so that you have a second parabola, this one passing through the origin and the point $(0,D)$ on the $y$-axis.  Picture a projectile $P$ moving along the first parabola and its mirror image $Q$ moving along the second.  We can think of $Q$ as a mass that is impervious to the normal form of gravity, but instead is subject to an identically strong "horizontal gravity" that pulls it toward the $y$-axis.  The physics is perfectly symmetric across the angle bisector of the $x$- and $y$-axes (i.e., the line $y=x$).
Because of the symmetry, the midpoint, call it $M$, between $P$ and $Q$ starts at the origin, travels out along the angle bisector to some maximum distance, and then heads back toward the origin, passing through it if you allow $P$ and $Q$ to continue past where they hit their respective axes.  It's easy to see (by drawing a picture, connecting the points $(D,0)$ and $(0,D)$ with a line of slope $-1$, and comparing that slope to the slopes of the parabolas at those points) that if the initial slope of the $P$-parabola is less than 1 (i.e., if you fire $P$ at an angle below 45 degrees), the maximum distance $M$ reaches occurs after the parabolas for $P$ and $Q$ pass through the axes (which is why I recommended continuing the $P$-parabola past the point of impact at $(D,0)$), that if the initial slope of the $P$-parabola is greater than 1, $M$'s maximum occurs before $P$ and $Q$ hit their axes, and that if the initial slope is exactly 1, $M$ reaches its maximum exactly when the projectiles $P$ and $Q$ hit their respective "grounds."
What's crucial here is that the distance $D$ that projectile $P$ travels before it hits the ground is never greater than what's allowed by the maximum distance that $M$ gets from the origin, with equality being achieved when $P$ is fired at 45 degrees.
But what is $M$?  Because it stays on the line $y=x$, it can be interpreted as a projectile fired straight out along that line subject to the vector sum of the two gravities, which is a force that points straight back along the same line.  Consequently, the maximum "height" (above the "$y=-x$" axis) achievable by $M$ is determined by its initial outward velocity, which is the component of the projectiles' ($P$ and $Q$) initial velocity along the line $y=x$.  And that component is obviously maximized when the entire velocity points in that direction.
So that takes care of the flat-ground case.  To go to the general case, it's helpful to rotate the flat-ground picture by 45 degrees, so that the axis of symmetry (i.e., the angle bisector) is now the $y$-axis.  The ground is now along the line $y=x$ and (normal) gravity points along the line $y=-x$.  The projectiles $P$ and $Q$ still travel along parabolas, but the line connecting them is now horiontal, and the midpoint projectile $M$ goes up and down the $y$-axis.  The maximum corresponds to a picture in which the two parabolas are tangent to the $y$-axis at the origin and to the horiontal line connecting them at the point of impact.
Now apply a linear transformation that takes $(x,y)$ to $(mx,y)$.  If $m\gt1$, this stretches out the angle between the direction of (normal) gravity and the direction of the ground, corresponding (if you re-rotate things) to sloping ground, but it leaves tangencies intact.  It has no effect on the $y$-axis (i.e., the angle bisector), and it stretches out the line connecting $P$ and $Q$ but keeps it horizontal.  Moreover, the parabolas are still parabolas -- you're really just using a different coordinate system to describe the same physics.  Case closed.
Footnote:  For lovers of conic sections, I heartily recommend the 1937 short Parabola, by experimental film maker Mary Ellen Bute, which highlights (literally!) sculptures by the artist Rutherford Boyd, whose work appeared regularly in the journal Scripta Mathematica.  It can be viewed in thumbnail size online  but for anyone interested it's available on disc 1 of Unseen Cinema: Early American Avant-Garde Film.