HIT IT! Hit the Target BRIEF ABSTRACT This projectile motion problem deals with the relationship between launch angle and starting height when attempting to hit a target. GENERAL INFORMATION FileName: Hitit Full title: Hit the Target Last Update: 6/3/96 Developers: Susan Clements, Mathematics Department, South Vigo High School, Terre Haute, IN 47802 USA Email: src@galileo.vigoco.k12.in.us Rosaline Secrest, Science Department, South Vigo High School, Terre Haute, IN 47802 USA Email: rls@galileo.vigoco.k12.in.us Aaron Klebanoff, Department of Mathematics, Rose-Hulman Institute of Technology, Terre Haute IN 47803 USA Contact: Aaron Klebanoff, Department of Mathematics, Rose-Hulman Institute of Technology, Terre Haute IN 47803 USA. Phone: 812-877-8151. Email: Aaron.Klebanoff@rose-hulman.edu. FAX: 812-877-3198. Support: The production of this material is supported by the National Science Foundation under Division of Undergraduate Education grant DUE-9352849: Development Site for Complex, Technology-Based Problems in Calculus with Applications in Science and Engineering and the Arvin Foundation of Columbus IN. STATEMENT OF PROBLEM For a science Olympiad project, you must build a device that can project an object over a horizontal distance of 450 feet. First, you need to determine the angle(s) required to project the object over the given desired distance with an initial velocity of 192 feet per second. Unfortunately, you won't know in advance whether the launch height, h, is at the same level as the target, although the height difference between the target and launch pad is guaranteed to be no more than a 10 feet. The steps below guide you through some preliminary considerations for building the device. (You will assume that air resistance is negligible and that the acceleration due to gravity is 32 feet per second-squared.) 1) Determine the launch angle(s) that will project the object exactly 450 feet down range (over level ground) with an initial velocity of 192 feet per second. (For now, take h = 0.) 2) Express the height of the target as a function of the launch angle, theta, and height, h. 3) Determine the range of launch angles (taking horizontal to be 0 degrees) that will allow the target to be within 10 feet of the launch height. 4) Determine and graph the rate of change of the target height with respect to the launch angle. Write a sentence or two discussing the implications of the result. KEYWORDS Projectile motion, acceleration due to gravity, horizontal and vertical position, horizontal and vertical velocity, rate of change, partial derivatives. TEACHER NOTES ISSUES RELATED TO THE PROBLEM Problem 3 could be done by guess and check. Problem 2 must be accomplished to complete problem 4. Prerequisites This problem is best given after the students are comfortable with anti-differentiation. The student must understand the relationship between acceleration, velocity, and position. The initial velocity given is at an angle, so it must be transformed to vertical and horizontal velocity. Although the problem entails partial differentiation, the students do not need the formalities to do it. It is, however, worth mentioning to them once they have completed the problem that they have used partial derivatives to get their result. Time allotment - time management Some time needs to be spent prior to this exercise breaking the initial velocity into its vertical and horizontal components. The student must also have a feeling for determining vertical and horizontal distances given the initial velocity. If these two aspects are already covered, the students, working in groups, should be able to complete the exercise in a one hour period. Expectations Although the solution to problem #2 seems straight forward, students may be confused over the two-variable dependence. Some students may not graph (or know how to) when it is appropriate. Once the graphs are done correctly, students will be able to explain to each other what the graphs mean. Future payoffs This exercise provides a sophisticated look at the classical projectile motion problem. This exercise provides a nice application for a derivative with respect to something OTHER than time. This exercise provides an introduction to multi-variable dependence. Extensions The horizontal distance in this problem was fixed, but some play could be made by changing this distance and relating it to the launch angle and the vertical position. The same could be done by playing with the initial velocity. Also, once the launch angle is calculated, the horizontal and vertical positions could be graphed parametrically with respect to time. And, for those who really like this problem, try revising the model to account for air resistance! POSSIBLE SOLUTION(S) 1) Determine the launch angle(s) that will project the object exactly 450 feet down range (over level ground) with an initial velocity of 192 feet per second. We first write the components of the velocity. In[2]:= Vx = 192 Cos[theta]; Vy = 192 Sin[theta] - 32 t; Next, we integrate to find the components of position. We let the parameter h stand for the launch height (measured in feet above the ground.) In[4]:= Px = Integrate[Vx, t]; Py = Integrate[Vy, t] + h; Now, by setting the horizontal position equal to 450 feet and the vertical position equal to 0 feet, we can find the angle, theta, and time T simultaneously. In[6]:= eqn1 = Px == 450; eqn2 = Py == 0; First we solve for the time using the horizontal position. In[8]:= Solve[eqn1, t] Out[8]= 75 Sec[theta] {{t -> -------------}} 32 In[9]:= T = t /. %[[1, 1]] Out[9]= 75 Sec[theta] ------------- 32 Now, we substitute the time into the vertical position and wish to know when that quantity is zero. We check first by plotting since we will be forced to solve numerically. In[10]:= SpecialPy = Py /. {h -> 0, t -> T} Out[10]= 2 -5625 Sec[theta] ----------------- + 450 Tan[theta] 64 In[11]:= Plot[SpecialPy, {theta, 0, Pi/2 - 0.18}] Out[11]= -Graphics- In[12]:= theta1 = theta /. FindRoot[SpecialPy == 0, {theta, 0.2}] Out[12]= 0.200655 In[13]:= theta2 = theta /. FindRoot[SpecialPy == 0, {theta, 1.4}] Out[13]= 1.37014 In[14]:= ThetaDeg1 = N[theta1 180/Pi] Out[14]= 11.4967 In[15]:= ThetaDeg2 = N[theta2 180/Pi] Out[15]= 78.5033 So, we find that either a 11.5 or 78.5 degree launch angle will do the job. 2) Express the height of the target as a function of the launch angle, theta, and height, h. In[16]:= Height[theta_] = Py /. t -> T Out[16]= 2 5625 Sec[theta] h - ---------------- + 450 Tan[theta] 64 3) Determine the range of launch angles (taking horizontal to be 0 degrees) that will allow the target to be within 10 feet of the launch height. We see immediately from a graph of Height = 0 in the h-theta plane that there are two launch angles for a given starting height. In[17]:= ImplicitPlot[Height[theta]==0, {h, -10, 10}, {theta, 0, Pi/2}, AspectRatio -> 1/GoldenRatio, PlotRange -> All, PlotLabel -> "From height h, angle theta: Hits 450 feet away", AxesLabel -> {"h [ft]", "theta [rad]"}, PlotPoints -> 50] Out[17]= -ContourGraphics- We first study the range on the small launch angle. In[18]:= ImplicitPlot[Height[theta]==0, {h, -10, 10}, {theta, 0.175, 0.225}, AspectRatio -> 1/GoldenRatio, PlotRange -> All, PlotLabel -> "From height h, angle theta: Hits 450 feet away", AxesLabel -> {"h [ft]", "theta [rad]"}, PlotPoints -> 50] Out[18]= -ContourGraphics- In[19]:= EqnH10 = Height[theta]==0 /. h -> 10 Out[19]= 2 5625 Sec[theta] 10 - ---------------- + 450 Tan[theta] == 0 64 In[20]:= ThetaMinH10 = theta /. FindRoot[EqnH10, {theta, 0.18}] Out[20]= 0.177487 In[21]:= EqnHN10 = Height[theta]==0 /. h -> -10 Out[21]= 2 5625 Sec[theta] -10 - ---------------- + 450 Tan[theta] == 0 64 In[22]:= ThetaMinHN10 = theta /. FindRoot[EqnHN10, {theta, 0.225}] Out[22]= 0.223842 In[23]:= ThetaMinH10 180/Pi //N ThetaMinHN10 180/Pi //N Out[23]= 10.1693 Out[24]= 12.8252 So, the launch angle decreases from 12.82 to 10.1 degrees as the launch height increases from -10 to 10 feet. Now, we study the range on the large launch angle. In[25]:= ImplicitPlot[Height[theta]==0, {h, -10, 10}, {theta, 1.368, 1.372}, AspectRatio -> 1/GoldenRatio, PlotRange -> All, PlotLabel -> "From height h, angle theta: Hits 450 feet away", AxesLabel -> {"h [ft]", "theta [rad]"}, PlotPoints -> 50] Out[25]= -ContourGraphics- In[26]:= ThetaMaxH10 = theta /. FindRoot[EqnH10, {theta, 1.371}] Out[26]= 1.37109 In[27]:= ThetaMaxHN10 = theta /. FindRoot[EqnHN10, {theta, 1.369}] Out[27]= 1.36917 In[28]:= ThetaMaxH10 180/Pi //N ThetaMaxHN10 180/Pi //N Out[28]= 78.5577 Out[29]= 78.4478 So, the launch angle increases from 78.45 to 78.55 degrees as the launch height increases from -10 to 10 feet. It is worth mentioning for a given launch height, the two launch angles to NOT sum to 90 degrees in general. See ISSUES IN SOLUTION. 4) Determine and graph the rate of change of the target height with respect to the launch angle. Write a sentence or two discussing the implications of the result. We note immediately that the change in the target height should be independent of the start height since a rise in X feet of the launch pad will translate to a rise in X feet of the target (since the Height is defined as the the vertical position when the projectile is 450 feet down range.) In[30]:= DerHeight = Height'[theta] Out[30]= 2 2 5625 Sec[theta] Tan[theta] 450 Sec[theta] - --------------------------- 32 It is interesting to note that there IS a launch angle (around 1.2 radians) for which small changes from that angle don't affect the height of the projectile. In[31]:= Plot[DerHeight, {theta, 0, 1.4}] Out[31]= -Graphics- In the range of launch angles that we are interested in, however, there is always a substantial change in Height with respect to launch angle. However, it is important to note that small changes in the higher launch angle are far more sensitive to height changes than the lower angle. This has practical implications: It is better to choose the lower launch angle (of the two choices) since small error in the launch angle won't have as much of an effect as if we had chosen the larger launch angle. For example, for the case when the the launch pad is 10 feet above the target (h = 10), we have a choice of two launch angles: 10.2 degrees or 78.5 degrees. Since a tiny error in choosing theta will greatly affect the target height, we choose the 10.2 degree launch angle to minimize the sensitivity. In[32]:= Plot[DerHeight, {theta, 0.175, 0.225}] Out[32]= -Graphics- In[33]:= Plot[DerHeight, {theta, 1.36, 1.37}] Out[33]= -Graphics- ISSUES IN SOLUTION Some might be quick to conclude that due to the lack of wind resistance, there are always two launch angles: call one alpha, then the other is 90 degrees - alpha (unless alpha = 45 degrees.) This isn't so (unless the launch height is the same as the target height, i.e. h = 0) as the graph below illustrates. The Graph In[34]:= ImplicitPlot[Height[theta]==0, {h, -500, 100}, {theta, 0, Pi/2}, AspectRatio -> 1/GoldenRatio, PlotRange -> All, PlotLabel -> "From height h, angle theta: Hits 450 feet away", AxesLabel -> {"h [ft]", "theta [rad]"}, PlotPoints -> 50] Out[34]= -ContourGraphics- In the special case when h = 0, the angles do some to 90 degrees. In[35]:= EqnH0 = Height[theta]==0 /. h -> 0; In[36]:= theta1 = theta /. FindRoot[EqnH0, {theta, 0.25}] Out[36]= 0.200655 In[37]:= theta2 = theta /. FindRoot[EqnH0, {theta, 1.5}] Out[37]= 1.37014 In[38]:= (theta1 + theta2) 180/Pi //N Out[38]= 90.