Angle Maximizing the Distance of a Projectile

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 $\frac{\phi}{2}$?).

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Consider what path is traced out by the projectile in the 2d velocity space (horizontal velocity x-axis; horizontal is "after rotating up so the ground is flat, gravity no longer vertical"). It starts somewhere on a circular arc, and thereafter follows a path 'down and to the right' at an angle $\phi$ to the vertical, at constant speed (corresponding to strength of g). Trace the line until it reaches the horizontal. This forms a triangle along with the origin.

The total distance travelled is just (total time in air)x(speed at max height), which is just proportional to the area of the triangle for fixed $\phi$. Possible initial angles give a family of triangles, with one side of fixed length and the opposite angle also fixed ($\pi/2 - \phi$); thus they fit in a circle and the maximum is clearly when the triangle is isosceles.

Here is a slightly more detailed solution along with a picture. This was written up by Davidac897 with the aid of Barry Cipra's writeup of ChrisJB's solution.

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 $\phi$. Therefore

In the velocity plane, the trajectory starts at $P$ 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 the negative of that of $P$ (this is true if we fire in a non-rotated frame, and the only difference here is that there is an extra force component in the $x$-direction).

In particular $\angle QOP = \theta' = \theta+\phi$, $\angle PQO = \frac{\pi}{2}-\phi$, and the segments intersect at $R$ in a right angle, where $\overline{QR}$ is the horizontal in the non-rotated frame. Then $OQ$ is the average horizontal speed of the projectile, and $OA$ is proportional to the total time in the area because it is half the total change in vertical velocity. Therefore, the area of the triangle, which is $\frac{(OQ)(OA)}{2}$, is proportional to the total distance traveled.

As remarked above, $OP$ is fixed, as is $\angle PQO$, and the area is maximized when $Q$ is the apex of an isosceles triangle, so $2 \angle QOP + \angle PQO = 2\theta + 2 \phi + \frac{\pi}{2}-\phi = \pi$, or $$\theta = \frac{\pi}{4}-\frac{\phi}{2}.$$

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With apologies for my obtuseness, it took me a while to see why the total distance traveled is proportional to the area of the triangle, so I've appended a somewhat longwinded version of your solution to my earlier answer. I really like the way working velocity space makes everything so easy! –  Barry Cipra Nov 23 '12 at 15:59
Thank you! This is beautiful! And exactly what I was looking for. I've written up the answer in a bit more detail. And like Barry Cipra, it took me a bit of time to figure out the details of the solution. –  David Corwin Nov 25 '12 at 4:58
@Davidac897, nice write-up -- and nice problem, too. I'll leave my own version posted just for the record. I see now that I failed to notice that $OQ$ represents the average velocity. (Incidentally, for some reason, I couldn't get the picture to load. Fortunately I didn't need it, since you used the same labeling I had used.) –  Barry Cipra Nov 25 '12 at 13:14

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 $v_y=v_0\sin\theta$ and $v_x=v_0\cos\theta$. Setting $v_0=g=1$ to clean out the clutter, we have

$$D=2\sin\theta\cos\theta = \sin(2\theta),$$

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 $g_y = g\cos\phi$, pointing down, and a horizontal component $g_x = g\sin\phi$, 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

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

We have $v_y = v_0\cos\theta'$ and $v_x = v_0\sin\theta'$. In this case it's convenient to adopt the clutter-cleaning convention $v_0 = \cos\phi$, which leaves us with

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

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

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

which this time is maximized, taking the value $1+\sin\phi$, when $2\theta+\phi = \pi/2$, 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 $v_0 = \cos\phi$, not $\sin\phi$. 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 $\phi\approx0$.) The full factor in $D$ that begs to be set equal to $1$ is $v_0^2/g\cos^2\phi$. If you stick with the convention $v_0=g=1$, you find that the maximum downhill distance, as a function of $\phi$ is

$$D_\max = \sec^2\phi+\sec\phi\tan\phi,$$

whose horizontal component is

$$H_\max = D_\max \cos\phi = \sec\phi + \tan\phi.$$

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 $P=(v_0\cos\theta', v_0\sin\theta')$ 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 $-v_0\sin\theta'$. (It's easy enough to work out the $v_x$ coordinates of $Q$ and $P'$, but it's unnecessary to do so.) Denoting points $A=(0,v_0\sin\theta')$ and $A'=(0,-v_0\sin\theta')$ 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 $\theta = \pi/4 - \phi/2$.

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Okay, I've simplified (I believe) what you wrote and added it to ChrisJB's answer, so you might or might not want to remove what's here. –  David Corwin Nov 25 '12 at 5:21

Here's another solution, which is certainly simpler than the first one I gave, and may also be simpler than ChrisJB's. It would again benefit from a drawing, but I hope the verbal description will suffice.

Let $V$ denote the (initial) velocity vector (of length $v_0$), and write $V=U+W$, where $U$ points in the "up" direction and $W$ points in the "downhill" direction. Together with the origin $O$, we have a parallelogram $OUVW$, with angle $\pi/2 + \phi$ at $O$. Let $u$ and $w$ be the lengths of $U$ and $W$.

The trick is to realize that the amount of time the projectile spends in the air is simply $T=2u/g$. Consequently, the downhill distance it travels is $D=wT=2uw/g$. The variable quantity here, $uw$, is proportional to the area of the parallelogram $OUVW$, which has fixed angles and a diagonal ($OV$) of fixed length. This area, and hence the distance $D$, is maximized when the parallogram is a rhombus, which is to say when $OV$ bisects the angle of $\pi/2 + \phi$ at $O$.

Added later: At the risk of complicating things, let me explain why $T=2u/g$.

Decompose the downhill vector $W$ into its horizontal and vertical components, writing $W=X-Y$, with components of (positive) length $x$ and $y$, respectively. (Note: the variables here denote velocities and speeds, not positions!) From the (absent, but easily drawn) picture, we have $\tan\phi = y/x$. Now, in addition to the projectile $P$ fired with initial velocity $V=U+W = U+X-Y$, imagine we fire two virtual projectiles: projectile $A$ straight up with velocity $U$ and projectile $B$ up and out with velocity $U+X$.

Because they have the same vertical component of velocity, $A$ and $B$ will always be at the same height; they simply separate horizontally with speed $x$. Similarly, $B$ and $P$, which have the same horizontal component of velocity, will always be vertically aligned, separating vertically at speed $y$. Gravity plays no role in the size, shape, or orientation of the triangle $\triangle ABP$; at any time $t$, it is a right triangle with a horizontal leg of length $|AB| = xt$ and vertical leg of length $|BP|=yt$. Hence the angle at $A$ is $\phi$, since the tangent is $yt/xt = y/x = \tan\phi$. This means that $AP$ is always parallel to the downhill slope, which means that projectiles $A$ and $P$ will hit the ground simultaneously.

But it's quite clear that $A$, being fired straight up with initial speed $u$, will land after time $T=2u/g$: It takes gravity time $u/g$ to bring the projectile to rest, and the same amount of time to bring it back down.

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It suffices to note that the distance-maximizing angle $\theta$ is linear in $\phi$. Since we know that $\theta=\pi/4$ for $\phi=0$, and since as $\phi\to -\pi/2$ (that is, as the incline gets steeper), we should have $\theta\to -\pi/2$ as well, we can determine the coefficients of the equation $\theta=a\phi+b$ by solving the system.

I think that it is intuitively obvious that $\theta$ is linear in $\phi$; to show this carefully one probably has to at least write down some of the physics in your question.

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That's not so obvious to me, though do you have any reasons why? Also, again, is there any physical intuition for what you're saying? –  David Corwin Jul 29 '10 at 21:34
Well there's good physical intuition for the limit statement (as the incline tends to vertical, you're sandwiched between $\phi$ and $-\pi/2$, as otherwise you're throwing backwards). As for linearity, I think you have to use something mathematical about the problem. My intuition is from working in a rotating frame with the origin at the center and the $y$-axis pointing towards the ball, but little in math is rigorous without writing down some math. –  Daniel Litt Jul 29 '10 at 22:08
Right, so I'm essentially looking for some physical or symmetric reason for linearity, which you propose requires some kind of equation manipulation. –  David Corwin Jul 29 '10 at 22:29

Here's a slightly longwinded solution that uses the symmetry of sinusoids. (Note that my $\theta$ is with respect to the horizontal, not the incline.)

For simplicity, let $g= v_0=1$. The vertical height $h(t)$ of the projectile above the incline is smaller than its $y$-coordinate by $x\tan\phi$, so $h(t)=t\sin\theta-t^2-t\cos\theta\tan\phi$. We know $h(0) = 0$, so the objective is to find the value of $\theta$ with the largest positive root of $h(t)$. (There is one positive root if we fire forward).

The graph of $h(t)$ is a parabola through the origin, so its positive root is twice the $t$-coordinate of its vertex. That puts the root at $x=\sin\theta-\cos\theta\tan\phi$. Let $R(\theta)$ be this $x$-value; for fixed $\phi$, $R(\theta)$ is a sinusoid.

For the $\theta$ values $\phi$ and $\pi/2$ (firing along the incline or vertically), $R(\theta)=0$, and for angles in between, $R(\theta)>0$. By the geometric properties of a sinusoid, the unique maximum is halfway between these angles. The optimal firing angle is therefore $(\phi+\pi/2)/2$, which is $\pi/4 - \phi/2$ up from the incline.

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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.

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.

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