physics - Angle Maximizing the Distance of a Projectile

Added 11/23/12: ChrisJB has given a really wonderful answer that avoids both derivatives and trigonemetric identities by considering geometry not in the (rotated) $xy$ plane but in the (rotated) $v_xv_y$ plane. It took me a while to understand why, in that answer, the distance traveled is proportional to the area of a triangle rather than a trapezoid, so I'm appending below my own original answer a slightly longwinded version of ChrisJB's, for the benefit of others who are as slow as I am. Credit (and upvotes) for it, however, should go entirely to ChrisJB. (You can skip now straight to the bottom.)

Here's something that at least avoids taking derivatives.

Let's start with a warm-up on flat ground. If you fire a projectile with vertical velocity $v_y$ and horizontal velocity $v_x$, the amount of time it spends in the air is $T=2v_y/g$ and the distance it travels is $D=v_xT$. As a function of firing angle $theta$, we have $v_y=v_0sintheta$ and $v_x=v_0costheta$. Setting $v_0=g=1$ to clean out the clutter, we have

$$D=2sinthetacostheta = sin(2theta),$$

which is clearly maximized, taking the value $1$, when $theta = pi/4$.

Now suppose the ground slopes down at angle $phi$. Let's rotate it up flat, and imagine firing at angle $theta' = theta + phi$. (We're not taking derivatives, so there should be no confusion in using the notation $theta'$.) The obvious problem is, gravity no longer points straight down. Instead it has a vertical component $g_y = gcosphi$, pointing down, and a horizontal component $g_x = gsinphi$, pointing to the right. In terms of their effect, the vertical component $g_y$ is a new (and reduced) gravity, while the horizontal component $g_x$ acts as a kind of additional magnetic force on the projectile, accelerating it in the $x$ direction. Thus the amount of time a projectile fired with vertical velocity $v_y$ spends in the air is $T=2v_y/g_y$, much as before, while the horizontal (actually downhill) distance it travels is now

$$D = v_xT + {1over2}g_xT^2 = 2v_y(v_xg_y + v_yg_x)/g_y^2.$$

We have $v_y = v_0costheta'$ and $v_x = v_0sintheta'$. In this case it's convenient to adopt the clutter-cleaning convention $v_0 = cosphi$, which leaves us with

$$D=2sintheta'(costheta'cosphi + sintheta'sinphi)=2sintheta'cos(theta'-phi),$$

using the angle addition formula $cos(x-y) = cos x cos y + sin x sin y$. The formula $2sin xcos y = sin(x+y)+sin(x-y)$ turns this into

$$D = sin(2theta'-phi) + sinphi = sin(2theta + phi) + sinphi,$$

which this time is maximized, taking the value $1+sinphi$, when $2theta+phi = pi/2$, which is to say, when $theta = pi/4 - phi/2$.

Added 11/15/12: Oops, I just fixed a minor mistake: the correct clutter-cleaning convention is $v_0 = cosphi$, not $sinphi$. I should have realized this right away from the fact that it needs to agree with the flat-ground convention, $v_0=1$, when $phi=0$. (I had elsewhere kept my sines and cosines straight using, for example, the fact that $g_x$ should be negligible for $phiapprox0$.) The full factor in $D$ that begs to be set equal to $1$ is $v_0^2/gcos^2phi$. If you stick with the convention $v_0=g=1$, you find that the maximum downhill distance, as a function of $phi$ is

$$D_max = sec^2phi+secphitanphi,$$

whose horizontal component is

$$H_max = D_max cosphi = secphi + tanphi.$$

Added 11/23/12: This is my longwinded version of ChrisJB's answer.

If you rotate the system so the ground is flat, you'll be firing at angle $theta' = theta + phi$ into a medium where gravity points down and to the right at angle $phi$. In the velocity plane, the trajectory starts at $P=(v_0costheta', v_0sintheta')$ and follows a straight line at angle $pi/2 - phi$ to the $v_x$ axis, through a point $Q$ on the $v_x$ axis, down to a point $P'$ with $v_y$ coordinate $-v_0sintheta'$. (It's easy enough to work out the $v_x$ coordinates of $Q$ and $P'$, but it's unnecessary to do so.) Denoting points $A=(0,v_0sintheta')$ and $A'=(0,-v_0sintheta')$ on the $v_y$ axis, we find that the total (downhill) distance traveled by the projectile is proportional to the area of the trapezoid $APP'A'$. (This is because, for a given $phi$, changes in velocity are proportional to changes in time.) If you draw the trapezoid, it's easy to see that its area is 4 times the area of the triangle $triangle OPQ$, $O=(0,0)$ being the origin. (This is the "easy to see" point that took me a while to see. If someone with the wherewithal to do so could insert an actual picture here, I would very much appreciate it.) The angle at $Q$ is fixed at $pi/2 - phi$ and the length of the side opposite $Q$ is fixed at $OP=v_0$. It doesn't require calculus to conclude that the triangle's area is maximized when $Q$ is the apex of an isosceles triangle, i.e., when $theta' = pi/4 + phi/2$, which translates back to $theta = pi/4 - phi/2$.