Projectile Motion: Finding the Optimal Launch Angle

1 Projectile Motion: Finding the Optimal Launch AngleNina HenelsmithWhitman CollegeMay 12, 20161 AbstractIf we want to throw a Projectile as far as possible, at what Angle should it be launched?This paper focuses on how the answer to this question changes depending on the look at launching projectiles onto di erently shaped hills, as well as how varying initialvelocity and height a ect the Launch Angle . Finally, we add air resistance to the projectileproblem and compare two di erent models: air resistance proportional to the Projectile svelocity and air resistance proportional to velocity you to Professor Russ Gordon for his helpful support and guidance for this you also to Karen Vezie, for her valuable edits and comments, and to ProfessorBarry Balof, for teaching the Senior Project math class and providing useful project Introduction42 The Projectile problem43 Equations of motion: no air resistance54 The Optimal Launch The distance function.

2 The enveloping parabola .. Derivation of the enveloping parabola: height maximization .. Derivation of the enveloping parabola: expanding circles .. The Projectile problem solution .. 125 Varying the impact The linear impact function .. The parabolic impact function .. The semicircular impact function .. The sinusoidal impact function .. Newton s method .. Varying initial conditions .. 216 Air Equations of motion: linear air resistance .. The level ground impact function .. The parabolic impact function .. Quadratic air resistance .. The numerical solution .. The drag coe cient.

3 347 Conclusion36 Index3831 IntroductionIn this paper, we examine how to find the Optimal Launch Angle , which is the angleat which a Projectile is launched that maximizes its horizontal distance traveled. We findthat this Angle depends on numerous factors, including the Projectile s initial velocity,the e ects of air resistance, and the surface upon which the Projectile lands. This paperaddresses these relationships in three parts: Finding a general solution for the optimallaunch Angle , exploring di erent landing surfaces, and adding the e ects of air the first four sections, we derive a solution to the Projectile problem by consideringthe Projectile s equations of motion.

4 We introduce the importance of the envelopingparabola and derive its equation in two ways. In the fifth section we examine specificexamples of the Projectile landing on di erently shaped hills, specifically linear, parabolic,semicircular and sinusoidal. We also explore the dependence of the Optimal Launch Angle onthe Projectile s initial height and velocity. Finally, in the sixth section we add air resistanceto the problem, examining both the linear and the quadratic model. We compare thesetwo models and aim to understand why certain models work better than others in a The Projectile problemWe define the Projectile problem as follows: a Projectile is launched from a tower ofheighth, with initial velocityv, and at an Angle measured with respect to the aim to find m, the Launch Angle that maximizes horizontal distance.

5 The projectileis launched onto a hill, which is defined by the function , called theimpact impact function varies depending on the shape of the surface we want to explore,and in this paper we will look at -functions that are linear, parabolic, semicircular andsinusoidal. Figure 1 shows a typical setup for the Projectile (x)p(x)hHorizontaldistanceVerticaldistan ceFigure 1: The Projectile Equations of motion: no air resistanceWe first consider the situation of a Projectile launched from a tower of heighthontosome impact function , ignoring the e ect of air resistance. In order to solve for m,weneed to find equations for motion in thex- andy- directions .

6 We define to be the angleabove the horizontal at which the Projectile is launched. The Projectile is launched with aninitial velocityv, which has magnitudev, and when broken up intox- andy-components,gives us the initial conditionsx(0) = 0;x0(0) =vcos ;y(0) =h;y0(0) =vsin .Without air resistance, acceleration in thex-direction is zero, while in they-direction it issolely due to gravity, whereg= m/s2. Thus we can solve the second-order di erentialequations to find our two motion equations. In each step, we integrate both sides of the5equation with respect totand apply our initial conditions. In thex-direction, we havex00(t) = 0;x0(t)=vcos ;x(t)=vtcos.

7 (1)The motion in they-direction is described byy00(t)= g;y0(t)= gt+vsin ;y(t)= 12gt2+vtsin +h.(2)We now have a set of parametric equations for the motion of the Projectile as a functionoft, but to maximize the Projectile s horizontal distance, we want to find a path function,p, that defines the Projectile s height as a function of horizontal distance,x. Solving fortin (1) and substituting into (2) yieldst=xvcos ,and thereforep(x)=h+vsin xvcos 12g xvcos 2=h+xtan gx22v2sec2 .(3)We now have one equation that describes the motion of the Projectile , which is useful infinding the Launch Angle that The Optimal Launch angleNext we will explore the process of Finding m, the Projectile s Optimal Launch an-gle.

8 This Angle will vary depending on the height and velocity at which the Projectile is6launched, as well as what type of surface the Projectile lands on. The impact function, ,is a function of horizontal distance,x. We find that for certain functions, obtaining aclosed-form solution for mis possible, while for other functions, we must instead turnto approximations of The distance functionFor each value of , there is a value ofxwhere the Projectile hits the hill, whichoccurs whenp(x)= (x). We call thisx-valued( ) since it varies depending on thelaunch Angle . By maximizingd( ), we can find the Angle that maximizes the Projectile shorizontal distance.

9 We note that for everyxiwithin the Projectile s horizontal range, wecan find a Launch Angle corresponding to at least one Projectile that hits thatxi. This fact,along with the Implicit Function Theorem (see [11]), lets us assumed( ) is a di erentiablefunction. With this condition, we can use implicit di erentiation on our distance function,p, to maximized( ) [3]. We have (d( )) =p(d( )) =h+d( ) tan d( )2g2v2sec2 .Di erentiation gives 0(d( ))d0( )=d( )sec2 +d0( ) tan gv2 d( )2sec (sec tan )+d( )d0( )sec2 .Sinced0( m) = 0, we find that0=d( m)sec2 m gv2d( m)2sec2 mtan md( m)=v2gcot m,which tells us that the maximum horizontal distance the Projectile travels is dependentupon initial velocity and gravity.

10 We now have a value for horizontal distance in termsof the Launch Angle . This value is independent of the impact function , but the anglethat maximizesxwill vary for di erent -functions, a connection we explore in Section The enveloping parabolaAnenveloping parabolais a path that encloses and intersects all possible projectilepaths. For each value of in [ , ], the enveloping parabola intersects the projectilepath at exactly one point, and at this point the two functions share a common tangentline (see [1]). Figure 2 shows the enveloping parabola that intersects each possible pathat exactly one point, where a possible path corresponds to a unique Launch Angle in therange [ , ].