OPTIMAL ANGLES FOR LAUNCHING PROJECTILES: LAGRANGE VS. CAS

Transcription

1 CANADIAN APPLIED MATHEMATICS QUARTERLY Volume 16, Number 3, Fall 2008 OPTIMAL ANGLES FOR LAUNCHING PROJECTILES: LAGRANGE VS. CAS ROBERT KANTROWITZ AND MICHAEL M. NEUMANN ABSTRACT. This article centers around the motion of a projectile that is launched from the top of a tower and lands on linear or parabolic mountains. The main goal is to provide explicit and manageable formulas for the angle of inclination that maximizes the distance the projectile travels. A general approach based on Lagrange multipliers is developed to obtain useful information not only for motion without air resistance, but also for the case when air resistance is proportional to the speed of the projectile. In this case, the optimal angle is expressed in terms of the Lambert W function. The explicit formulas for the optimal angle are employed to answer a number of natural questions. For instance, it turns out that the optimal angle is a decreasing function of the height of the launch tower. Also discussed is the extent to which computer algebra systems are helpful in this context. 1 Introduction and motivation The principal motivation for this article stems from the classical problem of finding the optimal angle for launching a projectile to maximize the distance that it travels. More specifically, here we consider the motion of a projectile that is launched from the top of a tower at time t = 0 with initial position 0, h, muzzle speed s, and angle of inclination θ with respect to the horizontal line y = h in the direction of the positive x-axis. The position of the projectile at time t 0 is denoted by xt, yt. We assume that the motion is subject to a constant force which, by Newton s law, results in an acceleration with components of magnitude g in the direction of the negative y-axis and k in the direction of the negative x-axis. The archetypical case arises when h = 0, k = 0, and g = 9.81 m/sec 2 is the gravitational constant. AMS subject classification: Primary 49K99; secondary 70B05. Keywords: Lagrange multipliers, projectile, ballistics, Lambert W function. Copyright c Applied Mathematics Institute, University of Alberta. 279

2 280 R. KANTROWITZ AND M. M. NEUMANN We first ignore air resistance and other possible factors that may influence the motion of the projectile. In this case, we are led to the initial value problem x t = k, x 0 = s cos θ, x0 = 0, y t = g, y 0 = s sin θ, y0 = h. Two elementary integration steps yield the familiar formulas xt = s cos θ t 1 2 k t2 and yt = h + s sin θ t 1 2 g t2, valid for all t 0 for which the projectile stays in the air until it hits a surface represented by the graph of a given function y = fx. The classical case of the motion of a projectile landing at ground level is described by f = 0, but we will also consider functions such as fx = mx and fx = mx 2 to model projectiles landing on linear and parabolic mountains. Our main interest is in comparing the trajectories of the projectile for different launch angles. To emphasize the dependence on θ, we often write xt, θ, yt, θ for the position of the projectile at time t when launched with the angle of elevation θ. We assume that, for each θ for which π/2 θ π/2, there exists some time t θ > 0 for which the impact yt θ, θ = f xt θ, θ occurs. A typical task is then to determine an angle θ for which the distance from the launch point 0, h to the point of impact xt θ, θ, yt θ, θ is maximal. One may also be interested in maximizing, for example, the arc length of the trajectory from 0, h to xt θ, θ, yt θ, θ or the distance from the origin to the point of impact. Of particular importance is the case of maximizing the lateral displacement xt θ, θ for π/2 θ π/2; in fact, some of the other cases may easily be reduced to this one. To present a uniform approach to this variety of optimization problems, we consider a function ρ of the two variables x and y that plays the role of a generalized distance function. The goal is then to maximize the function ρ xt, θ, yt, θ subject to the constraint yt, θ = fxt, θ. As indicated above, typical simple examples for ρx, y are x, x 2 + y 2, x 2 + y h 2, or the square roots of the last two expressions. We also consider the same kind of optimization problem for the case of motion under air resistance. In this context, we assume the retarding force to be proportional to the velocity vector with a negative constant

3 OPTIMAL ANGLES FOR LAUNCHING PROJECTILES 281 factor. This leads to the initial value problem x t = k αx t, x 0 = s cos θ, x0 = 0, y t = g αy t, y 0 = s sin θ, y0 = h, where α > 0 is the air resistance coefficient. It is well known and easily verified that the solution of this uncoupled pair of ordinary differential equations is given by and xt = k k + αs cos θ t + 1 e αt α α 2 yt = h g g + αs sin θ t + 1 e αt α α 2. In the special case when h = 0, k = 0, and f = 0, the corresponding motion of projectiles has been investigated, for instance, in [2, 3, 4, 6, 7, 9]. We mention in passing that, as α approaches zero, the preceding expressions for xt and yt tend to those describing the motion without air resistance. This follows by a straightforward application of l Hospital s rule. A natural strategy for attacking optimization problems of the indicated type is to solve the impact condition yt, θ = fxt, θ for t in terms of θ and thus to obtain a formula for the hit time t θ. Plugging the result into the distance function leads to the problem of maximizing dθ = ρ xt θ, θ, yt θ, θ for π/2 θ π/2. The final step then involves the solution of the critical point equation d θ = 0. Unfortunately, this approach meets intimidating obstacles even in relatively simple situations. For instance, when the task is to shoot a projectile as far as possible from a tower to ground level without air resistance and subject only to standard gravitation, then it is easily seen that the impact time is given by t θ = 1 s sin θ + 2gh + s g 2 sin 2 θ, which leads to the problem of maximizing the distance function 1 dθ = xt θ, θ = s cos θ t θ = s s g cos θ sin θ + 2gh + s 2 sin 2 θ.

4 282 R. KANTROWITZ AND M. M. NEUMANN While differentiation of d poses no problem, finding the solution of the critical point equation d θ = 0 is a challenging exercise unless h = 0. Computer algebra systems CAS such as Mathematica or Maple provide a solution in this and some other cases. However, as will be seen below, these systems do not always work, and, when they do, their solutions tend to be frighteningly complicated and provide little insight. In the next section, we employ the Lagrange multiplier theorem to develop a necessary condition for the optimal launch angle in an abstract setting that involves generalized distance and constraint functions ρ and f. It turns out that, in many instances, the optimal angle θ opt and the corresponding time of impact t opt are exactly the non-trivial solutions of the pair of equations x t t, θy θ t, θ = x θ t, θy t t, θ and yt, θ = fxt, θ. Our general result leads to surprisingly simple optimality conditions for the motion of projectiles with or without air resistance. In the third section, we specialize to the case when k = 0 and f = 0. Here we obtain explicit formulas for the optimal angle that allow us to answer several natural questions. For instance, in either the absence or presence of air resistance, it is shown that the optimal angle is a strictly decreasing function of the height h of the launch tower. Moreover, the optimal angle for motion without air resistance turns out to be the limit of the optimal angles for the case of air resistance as the resistance coefficient α tends to zero. For all this the Lambert W function proves to be an indispensable tool. In the last section, we establish and explore formulas for the optimal angle for projectiles launched to certain mountains. We also compare our approach with what Mathematica and Maple can do in this setting. 2 The Lagrange multiplier approach Throughout this section, we consider continuously differentiable functions f : J R on an open interval J in R and ρ : Ω R on an open subset Ω of R 2. Let σx = ρx, fx for all x J for which x, fx Ω. Also, let x, y : Γ R be continuously differentiable functions on an open subset Γ of R 2 such that xt, θ, yt, θ Ω for all t, θ Γ. In this general setting, we obtain the following necessary condition for the solution of a canonical constrained optimization problem. Theorem 1. Let t, θ Γ be a point for which xt, θ J, and

5 OPTIMAL ANGLES FOR LAUNCHING PROJECTILES 283 suppose that t, θ provides a local extremum of the function ρxt, θ, yt, θ subject to the constraint yt, θ = fxt, θ. Then either xt, θ is a critical point of the function σ, or 2 x t t, θ y θ t, θ = x θ t, θ y t t, θ. Proof. Let x, y = xt, θ, yt, θ, and consider the functions u and v given by ut, θ = yt, θ fxt, θ and vt, θ = ρxt, θ, yt, θ. We separate two possibilities for the point t, θ. If ut, θ is the zero vector, then, by the chain rule, we have y t t, θ = f x x t t, θ and y θ t, θ = f x x θ t, θ and therefore, x θ t, θ y t t, θ = f x x t t, θ x θ t, θ = x t t, θ y θ t, θ, as desired. In the remaining case where ut, θ is non-zero, we employ the fact that t, θ yields a local extremum of vt, θ subject to the constraint ut, θ = 0. Hence the Lagrange multiplier theorem yields the existence of some λ R for which vt, θ = λ ut, θ. By the multivariate chain rule, this identity translates into the following two equations: ρ x x, y x t t, θ + ρ y x, y y t t, θ = λ y t t, θ f x x t t, θ, ρ x x, y x θ t, θ + ρ y x, y y θ t, θ = λ y θ t, θ f x x θ t, θ. Cross-multiplication and simplification then leads to the identity λ ρ x x, y x t t, θ y θ t, θ ρ y x, y f x x θ t, θ y t t, θ = λ ρ x x, y x θ t, θ y t t, θ ρ y x, y f x x t t, θ y θ t, θ

6 284 R. KANTROWITZ AND M. M. NEUMANN which may be rewritten in the form λρ x x, y + ρ y x, y f x x t t, θ y θ t, θ x θ t, θ y t t, θ = 0. If the last of the three main factors of the preceding product is zero, then we are finished. Otherwise, we conclude that λ = 0 or ρ x x, y + ρ y x, y f x = 0. If λ = 0, then it easily follows from vt, θ = λ ut, θ = 0, 0 and x t t, θ y θ t, θ x θ t, θ y t t, θ that ρ x x, y = ρ y x, y = 0. Thus in either of the two remaining possibilities we have ρ x x, y + ρ y x, y f x = 0. But this means precisely that σ x = 0 and hence completes the argument. We now apply Theorem 1 to the two cases of projectile motion considered in the introductory section. Corollary 2. For given constants g, k, h and s with s 0, consider xt, θ = s cos θ t 1 2 k t2 and yt, θ = h + s sin θ t 1 2 g t2. Let t > 0 and θ R be such that xt, θ J and xt, θ, yt, θ Ω, and suppose that t, θ provides a local extremum of the function ρxt, θ, yt, θ subject to the constraint yt, θ = fxt, θ. Then either xt, θ is a critical point of σ, or and θ solves the equation t = s k cos θ + g sin θ s 2 sin θ h + k cos θ + g sin θ gs 2 2 k cos θ + g sin θ 2 s 2 cos θ = f k cos θ + g sin θ ks 2 2 k cos θ + g sin θ 2. Proof. The result is immediate from 2 and the constraint yt, θ = fxt, θ. Indeed, under the present conditions, the identity x t t, θy θ t, θ = x θ t, θy t t, θ

7 OPTIMAL ANGLES FOR LAUNCHING PROJECTILES 285 from Theorem 1 means precisely that s cos θ kt s cos θt = s sin θ gt s sin θt which reduces to st = g sin θ + k cos θ t 2. The last equality ensures that g sin θ +k cos θ 0 whenever s, t 0 and hence leads to the desired formulas for t and θ. For given constants g, k, h, s, and α with s, α 0, con- Corollary 3. sider xt, θ = k α t + k + αs cos θ α 2 1 e αt and yt, θ = h g α Let t > 0 and θ R be such that t + g + αs sin θ α 2 1 e αt. xt, θ J and xt, θ, yt, θ Ω, and suppose that t, θ provides a local extremum of the function ρxt, θ, yt, θ subject to the constraint yt, θ = fxt, θ. Then either xt, θ is a critical point of σ, or t = 1 α ln 1 + and θ solves the equation h g 1 α 2 ln αs + k cos θ + g sin θ = f kα 2 ln 1 + αs k cos θ + g sin θ + αs k cos θ + g sin θ s g + αs sin θ α αs + k cos θ + g sin θ + s k + αs cos θ α αs + k cos θ + g sin θ.

8 286 R. KANTROWITZ AND M. M. NEUMANN Proof. For the present choice of xt, θ and yt, θ, condition 2 for the partial derivatives from Theorem 1 assumes the form k α k + αs cos θ s cos θ + e αt 1 e αt α α = s sin θ α 1 e αt g α When t > 0 this identity may be simplified to g + αs sin θ + e αt. α k cos θ + k cos θ + αs cos 2 θ e αt = g sin θ g sin θ + αs sin 2 θ e αt or, equivalently, αs + k cos θ + g sin θ e αt = k cos θ + g sin θ. Since the condition that s, α 0 ensures that both αs + k cos θ + g sin θ and k cos θ + g sin θ are non-zero, the last identity may be rewritten as t = 1 α ln 1 + αs. k cos θ + g sin θ This establishes the assertion for t, and the final formula for θ is now immediate from the identity yt, θ = fxt, θ. We note in passing that, for fixed g, k, s, and θ, a simple application of l Hospital s rule shows that 1 lim α 0 + α ln 1 + αs = k cos θ + g sin θ s k cos θ + g sin θ. In this sense, the optimality condition for motion without air resistance from Corollary 2 is obtained as the limit of the corresponding condition for motion under air resistance from Corollary 3 as the resistance coefficient α tends to 0. Consequently, it seems reasonable to expect that the optimal angle for launching a projectile without air resistance is obtained as the limit of the optimal angles for motion under air resistance when α approaches 0. This issue will be addressed in Corollary 6 below. To illustrate the role of the critical point condition for σ in the preceding results, we consider the following two examples. If ρx, y = x, then this condition becomes vacuous, since σ x = 1 for all x. On the

9 OPTIMAL ANGLES FOR LAUNCHING PROJECTILES 287 other hand, if ρx, y = x 2 +y h 2 1/2, then a point x, y 0, h on the graph of f satisfies σ x = 0 precisely when x + y hf x = 0 or, equivalently, f x = x/h y. Geometrically, this condition means that the line through the launch point 0, h and the point of impact x, y is perpendicular to the line tangent to the graph of f at x, y. In particular, if f 0 = 0, then x = 0 is a critical point of σ that corresponds to the launch angle θ = π/2. As will be seen in the last section, this may well be the optimal angle to maximize the distance from 0, h to the point of impact. 3 Launching from a tower to ground level In this section, we discuss our optimal angle problem in the special case where k = 0, f = 0, and ρx, y = x. Clearly, the corresponding function σx = x has no critical point. We note that the choice ρx, y = x 2 + h 2 would lead to exactly the same results, since here the only critical point of σ yields the minimal distance to 0, h. Thus, not surprisingly, the optimal angle in the setting of the next two results also maximizes the distance from the launch point 0, h to the point of impact. Theorem 4. For given constants g, s > 0 and h 0, let xt, θ = s cos θ t and yt, θ = h + s sin θ t 1 2 g t2. Then there exists a unique point t opt, θ opt that maximizes the function xt, θ subject to the constraint yt, θ = 0 for t 0 and π/2 θ π/2. In fact, 1 3 θ opt = arcsin and t opt = 1 2s2 + 2gh, 2 + 2gh/s 2 g while the maximal distance is given by xt opt, θ opt = s g s2 + 2gh. Proof. Evidently, the distance function d considered in 1 assumes its maximum at some point θ for which π/2 < θ < π/2, and the corresponding impact time t satisfies t > 0. Because k = 0 and f = 0, the last equation of Corollary 2 simplifies to h + s2 g s 2 2g sin 2 θ = 0.

10 288 R. KANTROWITZ AND M. M. NEUMANN But this means precisely that sin θ = 1/ 2 + 2gh/s 2 and hence leads to the assertion regarding θ opt. Moreover, again by Corollary 2, we have t opt = s = 1 2s2 + 2gh. g sin θ opt g The final formula for the maximal distance is now immediate. The formula for θ opt of the preceding result was found earlier, in Chapter 2 of [8], based on Mathematica. Actually, the version of Mathematica available at that time did not handle the problem until the second author implemented the substitution u = sin θ to eliminate the trigonometric functions in the critical point equation for the distance function d introduced in 1. Version 6.0 of Mathematica provides directly the solution of this critical point equation, but in an equivalent form involving arccos. Formula 3 turns out to be more useful, since it makes transparent that θ opt is a strictly decreasing function of the single parameter µ = gh/s 2. In particular, it follows that, for fixed g and s, the optimal angle is a strictly decreasing function of the height h of the launch tower. Plausible as this result may be, we do not know how to prove it without an explicit formula for θ opt. The reader is invited to find an elementary argument using only basic physics. Moreover, 3 confirms the expectation that θ opt tends to 0 as h and to π/4 as h 0. In the same vein, it follows that θ opt is increasing in s and decreasing in g when the other parameters are kept fixed. Chau [1] and Groetsch [5] consider a similar problem for launching a projectile from ground level to level h, implicitly assuming, of course, that the initial speed is sufficient to reach this level. We note that Theorem 4 remains valid for negative h and then covers the setting considered in [1] and [5]. Chau [1] employs a critical point approach for a suitable function of two variables, while Groetsch [5] uses the Lagrange multiplier theorem in this particular case. His approach inspired some of the results of our paper. To establish the counterpart of Theorem 4 for the case of air resistance, more machinery is needed. By basic calculus, the function T given by T w = w e w for all w 1 is strictly increasing and maps the interval [ 1, onto [ 1/e,. The inverse of T is known as the Lambert W function. It maps [ 1/e, onto [ 1, and has become a useful tool for the solution of equations involving the exponential function; for more details, history, and references, we refer to [9]. In Mathematica the notation for the Lambert W function is ProductLog, while Maple

11 OPTIMAL ANGLES FOR LAUNCHING PROJECTILES 289 uses just LambertW. We will employ the fact that W u > 0 for all u > 1/e, which is immediate from T w > 0 for all w > 1. Theorem 5. For given constants g, s, α > 0 and h 0, let xt, θ = s cos θ α 1 e αt and yt, θ = h g α t + g + αs sin θ α 2 1 e αt. Then there exists a unique point t opt, θ opt that maximizes the function xt, θ subject to the constraint yt, θ = 0 for t 0 and π/2 θ π/2. In fact, with the notation γ = αs/g the formula γ W γ 2 1 e 1 α2 h/g 4 θ opt = arcsin γ 2 1 W γ 2 1 e 1 α2 h/g holds if γ 1, whereas if γ = 1. Moreover, 1 θ opt = arcsin e 1+α2 h/g 1 t opt = 1 1 α ln αs + g sin θ opt while the maximal distance is given by xt opt, θ opt = = s g + αs sin θ opt g αs + g sin θ opt + αh g, s2 cos θ opt αs + g sin θ opt. Proof. We first observe that the condition yt, θ = 0 entails that t αh + s /g + 1/α. Thus it is immediate that the set D of all points t, θ for which t 0, π/2 θ π/2, and yt, θ = 0 is bounded, closed, and non-empty. Hence, by continuity, there exists a point t, θ in D for which xt, θ is maximal, and it is clear that this occurs when t > 0 and π/2 < θ < π/2. Consequently, it remains to solve the last

12 290 R. KANTROWITZ AND M. M. NEUMANN equation of Corollary 3 for sin θ in the special case when k = 0 and f = 0. Evidently, sin θ > 0 in the maximal case. Isolating the logarithmic term, we obtain the identity 5 ln 1 + αs g sin θ = αs g + αs sin θ g αs + g sin θ + α2 h g. We now apply the exponential function and collect the terms involving sin θ on one side of the resulting equation so that g sin θ αs g + αs sin θ αs + g sin θ exp = exp α2 h. g αs + g sin θ g To solve for sin θ in terms of the Lambert W function, we note that the preceding equation is of the form p e q = r, where r does not depend on sin θ. Since W ue u = u for all u 1, our strategy is to find c 0 and d both independent of sin θ so that cp = d + q. Indeed, for such c and d we would arrive at cpe cp = cpe d+q = ce d r and hence cp = W cpe cp = W ce d r provided that cp 1. This would then lead to a simple formula for sin θ, since the right-hand side is independent of θ. In the present setting, the condition cp = d + q means that cg sin θ αs g + αs sin θ = d + αs + g sin θ g αs + g sin θ which may be simplified to cg 2 sin θ = dg αs + g sin θ + αs g + αs sin θ. Collecting terms involving sin θ, we may rewrite this equation in the form cg 2 dg 2 α 2 s 2 sin θ = d + 1gαs. Evidently, the equation holds with the choice d = 1 and c = α2 s 2 g 2 1 = γ 2 1, since both sides are equal to zero. Because cp = d + q d = 1, it follows that γ 2 1 g sin θ = c p = W ce d r = W γ 2 1e 1 α2 h/g. αs + g sin θ

13 OPTIMAL ANGLES FOR LAUNCHING PROJECTILES 291 Note that c 0 precisely when γ 1, and, in this case, the preceding formula easily leads to a formula for sin θ and hence the optimal angle itself. In fact, we obtain γ 2 1 g sin θ = αs + g sin θ W γ 2 1e 1 α2 h/g and therefore, sin θ = γw γ 2 1e 1 α2 h/g γ 2 1 W γ 2 1e 1 α2 h/g. This shows that the optimal angle is unique and given by 4 when γ 1. In the remaining case γ = 1, equation 5 assumes the simple form ln = 1 + α2 h sin θ g which has the unique solution sin θ = 1/e 1+α2 h/g 1, as claimed. The final assertions are now immediate from 5 and the formula for t provided in Corollary 3. Since Mathematica fails to solve equation 5 even when assisted with the substitution u = sin θ, it is perhaps not surprising that the preceding proof requires some work. On the other hand, version 11 of Maple handles 5 nicely, but ignores the case γ = 1. We mention that our result for the optimal angle reduces to the one provided by Packel and Yuen in Theorem 3 of [9] for h = 0. Our use of the Lambert W function was inspired by [9], but our approach appears to be somewhat shorter even in this particular case. Using the data s = 330 m/sec, h = 50 m, and g = 9.81 m/sec 2, we illustrate in Figure 1 the paths of projectiles launched at angles θ = π/6 dotted, θ = π/15 dashed, and θ opt = solid, all subject to a coefficient of drag α = 0.2 sec 1. As Groetsch aptly remarks in [6], resistance is the enemy of symmetry, and here, of course, the paths of the projectiles launched from a tower are not parabolic as they would be without air resistance.

14 292 R. KANTROWITZ AND M. M. NEUMANN We now investigate the optimal angle of Theorem 5 as a function of h when all the other parameters are fixed. For this the closed form representation 4 of the optimal angle in terms of the Lambert W function turns out to be crucial. Clearly, the continuity of W ensures that θ opt depends continuously on h and hence approaches the optimal angle for launching from ground level as h 0. Similarly, because W 0 = 0, it follows from Theorem 5 that the optimal angle tends to zero when h. Moreover, it turns out that θ opt is a strictly decreasing function of h. Indeed, this is again a consequence of the formulas for θ opt provided in Theorem 5. The case γ = 1 is obvious, and, when γ 1, a routine computation based on 4 shows that θ opth = α2 γγ W ue 1 α2 h/g gγ 2 1 W u 2 cos θ opt h where u = γ 2 1e 1 α2 h/g. This ensures that θ opth < 0 for all h 0 simply because W u > 0, while a closed form for W u is not needed here. It seems it would be difficult to establish that θ opt decreases as a function of h without explicit formulas such as the ones provided in Theorem 5. As a by-product of our monotonicity result together with Theorem 2 of [9], we obtain that θ opt h < θ opt 0 < π/4 for all h > 0. In the special case h = 0, Packel and Yuen [9] proved that the optimal angle is a decreasing function of the parameter γ. The formulas for θ opt from Theorem 5 reveal that this result ceases to be true for h > 0. In fact, in this case, θ opt even fails to be a function of γ alone. It remains open if θ opt is decreasing in α, although this happened to be the case

15 OPTIMAL ANGLES FOR LAUNCHING PROJECTILES 293 in all numerical examples we checked. We also leave it to the reader to explore how θ opt depends on g when the other parameters are kept fixed. To conclude this section, we show that the results of Theorems 4 and 5 are connected by taking the limit as air resistance approaches 0. Corollary 6 supports with a rigorous proof what may be suggested by our physical intuition. Remarkably, Mathematica handles the limit of the next result even without the need for substitutions or other shortcuts. Maple, however, is unable to perform this computation. Corollary 6. Given g, s > 0 and h 0, let θ opt α denote the optimal angle provided by Theorem 5 for the air resistance coefficient α > 0, and let θ opt be the optimal angle provided by Theorem 4 for the case of no air resistance. Then θ opt α θ opt as α 0 +. Proof. By 3 and 4, it remains to establish the convergence γα W γα 2 1 e 1 α2 h/g γα 2 1 W γα 2 1 e 1 α2 h/g gh/s 2 as α 0 +, where γα = αs/g. Note that the quotient on the left-hand side may be rewritten in the form W γα 2 1 e 1 α2 h/g γα 1 + W γα 2 1 e 1 α2 h/g /γα and that the resulting new numerator approaches W 1/e = 1 as α 0 +. Hence our task reduces to proving that 1 + W γα 2 1 e 1 α2 h/g γα 2 + 2gh/s 2 as α 0 +. Fortunately, a closely related case was settled in [9, p. 348] where a suitable power series was employed to establish the convergence 1 + W β 2 1 e 1 β 2 as β 0 +. To take advantage of this result, we introduce βα = 1 + γα 2 1 e α2 h/g for all α > 0.

16 294 R. KANTROWITZ AND M. M. NEUMANN These quantities were born to satisfy βα 2 1 e 1 = γα 2 1 e 1 α2 h/g, and it is clear that βα 0 + as α 0 +. Because 1 + W γα 2 1 e 1 α2 h/g = 1 + W βα 2 1 e 1 βα γα βα γα for all α > 0, it remains to be seen that βα/γα tends to 1 + gh/s 2 as α 0 +. But this is not difficult. Indeed, by l Hospital s rule, we obtain βα 2 lim α 0 + γα 2 = lim g 2 + α 2 s 2 g 2 e α2 h/g α 0 + α 2 s 2 2αs 2 α 2 s 2 g 2 2αh/g e α2 h/g = lim α 0 + = lim α 0 + the desired convergence. = 1 + gh s 2, 2αs 2 1 α 2 s 2 g 2 h gs 2 e α2 h/g 4 Launching projectiles to linear and parabolic mountains In this last section, we revisit the motion of projectiles without air resistance, but now for certain non-trivial choices of the impact function y = fx. Unless noted otherwise, the goal is to maximize the canonical distance function ρx, y = x. First let k = 0. In this case, the optimality condition of Corollary 2 simplifies to h + s2 g s 2 2g sin 2 θ = f s 2 g cot θ. A particularly powerful substitution for this formula is u = cot θ. Indeed, because of the identity csc 2 θ = 1 + cot 2 θ, the preceding equation turns into 6 2gh s u2 = 2g s 2 s 2 f g u. We now solve this optimality condition in a few typical cases.

17 OPTIMAL ANGLES FOR LAUNCHING PROJECTILES First let fx = mx with some arbitrary constant m. Then, as in Theorem 4, our optimization problem certainly has a solution. To find a formula for the optimal angle, we note that 6 reduces to the simple quadratic equation u 2 + 2mu = 1 + 2µ with our old friend µ = gh/s 2. Since the maximal solution occurs for 0 < θ < π/2, we have u > 0 and therefore θ opt = arccot u = arccot m + m µ. Evidently, the result agrees with the one found in 3 for the case m = 0. Here, somewhat surprisingly, the optimal angle turns out to be a function of only m and µ. In particular, our solution formula makes it obvious that θ opt increases in s and decreases in g and h when all the others parameters are kept fixed. Moreover, since the function ψ given by ψm = m+m µ 1/2 satisfies ψ m < 0 for all m, it follows that θ opt is an increasing function of the slope m. The present problem was also considered in [8]. Specifically, the second author employed Mathematica to solve the impact condition yt, θ = mxt, θ for t in terms of θ and then to find the critical points of the resulting distance function. The version of Mathematica used in Sections 4.10 and 7.4 of [8] led to θ opt = arcsin 2ghs 2 + 2s 4 + 2m 2 s 4 ± 2 2ghm 2 s 6 + m 2 s 8 + m 4 s 8 1/2 4g 2 h 2 + 8ghs 2 + 4s 4 + 4m 2 s 4, which makes it hard see that θ opt is decreasing, for instance, in h. Moreover, the solution provided by the latest version of Mathematica is equally intimidating and involves eight possibilities for θ opt based on arccos. Each of them has the same complexity as the one provided above and does not improve even when the FullSimplify command is employed. Of course, the blame here lies not with Mathematica, but rather with the critical point method which appears to provide significantly less insight than our Lagrange approach from Corollary 2. We note that Mathematica handles 6 well, but fails to work without the u = cot θ substitution. Maple performs comparably to Mathematica for this and similar problems. If one replaces the distance function ρx, y = x by either x 2 + y 2 or x 2 +y h 2, then the optimal angle remains the same. Indeed, for every point x, y on the graph of f, the condition σ x = 0 holds precisely when ρx, y is minimal.

18 296 R. KANTROWITZ AND M. M. NEUMANN 4.2 Now let fx = mx 2, where m is a real constant for which g + 2ms 2 > 0. This condition ensures that, for each θ for which π/2 θ π/2, there exists some t 0 for which the impact equation yt, θ = mxt, θ 2 holds and that our optimization problem for ρx, y = x has a maximal solution. Since 6 simplifies to 1 + 2ms2 g u 2 = 1 + 2gh s 2, the optimal launch angle is given by the formula 1 + 2gh/s θ opt = arccot u = arccot ms 2 /g = arccot gs 2 + 2g 2 h gs 2 + 2ms 4. Thus θ opt is decreasing in h and g, but increasing in s and m, when all the other parameters are fixed. The critical point approach to this problem was discussed in Sections 4.11 and 7.5 of [8]. For this method, the current version of Mathematica leads to 2g 2 h + gs 2 θ opt = ± arccos ± 2g 2 h + 2gs 2 + 2ms 4. Except for the issue of sign, this is equivalent to the preceding solution, but, as in 4.1, the result appears to be less useful. Maple yields a similar formula in terms of arctan. While the optimal angle remains the same when ρx, y = x 2 + y 2, the situation changes for the choice ρx, y = x 2 + y h 2. To construct a specific example, suppose that h > 0 is large enough to satisfy s s2 + 2gh h g 2, and consider fx = mx 2 with m = 1/h. Then it is easily seen that σx = ρx, fx < h 2 = σ0 for all 0 < x h/2. Since the last formula of Theorem 4 and the choice of h ensure that xt, θ h/2 whenever yt, θ = fxt, θ, we conclude that here the critical point x = 0 of σ yields the optimal solution and hence that θ opt = π/2 is the unique launch angle that maximizes the distance from 0, h to the point of impact. This example shows that it is essential to include the critical points of σ in Theorem 1.

19 OPTIMAL ANGLES FOR LAUNCHING PROJECTILES We finally explore the optimality condition of Corollary 2 in the case where k may be non-trivial, while f = 0. In this case, the condition for θ assumes the form h + s 2 sin θ k cos θ + g sin θ g s 2 2 k cos θ + g sin θ 2 = 0. We do not recommend solving this equation by hand. After the substitution u = sin θ and a simplification command both Mathematica and Maple provide the solution θ opt = arcsin K where K = ± g 2 + k 2 2h 2 k 2 + ghs 2 + s 4 ± k 2 g 2 + k 2 s 2 2gh + s 2 3 2g 2 + k 2 h 2 g 2 + k 2 + 2ghs 2 + s 4. Complicated as this formula may be, it is much simpler than the long solution that results from Mathematica through a critical point approach; see Section 7.4 of [8]. Figure y 100 Β h Β h P Β x 100 Moreover, the appearance of the term g 2 + k 2 suggests some simplification. Indeed, as indicated in Figure 2, let P denote the point of

20 298 R. KANTROWITZ AND M. M. NEUMANN intersection of the x-axis and the line through 0, h with direction vector k, g. Clockwise rotation around P by the angle β = arctank/g then yields the setting of 4.1 with the new parameters ĝ = g 2 + k 2, ĥ = h sec β = h g g2 + k 2, and m = tan β = k g. Since the optimal angle ˆθ opt for the rotated setting also maximizes the distance from P to the point of impact, we obtain ˆθ opt = θ opt + β. Thus the main result of 4.1 leads to the surprisingly simple formula θ opt = arccot k g + k 2 g hg2 + k 2 gs 2 arctan k. g Again, the message is that, even with all the power of a CAS, the approach matters. REFERENCES 1. W. Chau, Optimal initial angle to fire a projectile, Pi Mu Epsilon J , N. de Mestre, The Mathematics of Projectiles in Sport, Cambridge University Press, Cambridge, C. W. Groetsch, Tartaglia s inverse problem in a resistive medium, Amer. Math. Monthly , C. W. Groetsch, Inverse Problems, Math. Assoc. of America, Washington, DC, C. W. Groetsch, Geometrical aspects of an optimal trajectory, Pi Mu Epsilon J , C. W. Groetsch, Another broken symmetry, College Math. J , C. W. Groetsch and B. Cipra, Halley s comment projectiles with linear resistance, Math. Magazine , M. M. Neumann and T. L. Miller, Mathematica Projects for Vector Calculus, Kendall-Hunt Publishing Company, Dubuque, IA, E. W. Packel and D. S. Yuen, Projectile motion with resistance and the Lambert W function, College Math. J ,

21 OPTIMAL ANGLES FOR LAUNCHING PROJECTILES 299 Corresponding author: Michael M. Neumann, Department of Mathematics and Statistics, Mississippi State University, Mississippi State, MS 39762, USA address: Robert Kantrowitz, Department of Mathematics, Hamilton College, Clinton, NY 13323, USA address:

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