Setting up the cutoff for baseball relay BRIEF ABSTRACT This uses a simple differential equation to model the relay between an outfielder and an infielder throwing the ball home. GENERAL INFORMATION FileName: RELAY Full title: Setting up the Cutoff for Baseball Relay Developer: Brian J. Winkel, Department of Mathematics, Rose-Hulman Institute of Technology, Terre Haute IN 47803 USA. Contact: Brian J. Winkel, Department of Mathematics, Rose-Hulman Institute of Technology, Terre Haute IN 47803 USA. Phone: 812-877-8412. Email: winkel@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 A baseball player (outfielder) throws a ball with initial velocity 100 ft/sec toward home plate some 400 feet away. Another player (an infielder who cuts the ball off - catches it - from the outfielder's throw) places himself in direct line between the outfielder and home plate and catches the ball x feet from home plate and then throws (relays) the ball home with initial velocity 110 ft/sec. It takes 2 seconds to accomplish the relay. Assume that due to air resistance the velocity v = v(t) obeys the differential equation dv/dt = -.1 v and ignore vertical motion. (a) Find the time, T, it takes for the ball to reach home as a function of x. (b)Find x which minimizes T. (c) For a range of infielder capabilities, i.e., initial throwing velocities from 60 ft/sec to 140 ft/sec, prescribe the cutoff distance x and the total time for the relay to the plate. (d) Consider the infielder who can throw 110 ft/sec. Would you rather improve the infielder's relay time, say down to 1.8 sec, or increase his initial velocity on the ball by 10 ft/sec? Explain. KEYWORDS Projectile motion, first-order differential equation. TEACHER NOTES ISSUES RELATED TO THE PROBLEM Prerequisites Student should be able to solve separable first order differential equation and apply antilogarithms - or know how to make a computer/calculator do it. Time allotment - time management Expectations Future payoffs Extensions References and Sources Initial problem concept and wording for (a) and (b) from Roland Minton, Department of Mathematics, Roanoke College, Salem VA 24153 USA. POSSIBLE SOLUTION(S) (a) Find the time, T, it takes for the ball to reach home as a function of x. We model the outfielder's throw with initial velocity of 100 ft/sec and initial position of 400 feet from the plate. We use the following convention: distance toward the plate is negative and velocity toward the plate is also negative. solv = DSolve[{v'[t] == -.05 v[t],v[0] == -100},v[t],t] -100 {{v[t] -> -------}} 0.05 t E vo[t_] = v[t]/.solv[[1]]; sols = DSolve[{s'[t] == vo[t],s[0] == 400},s[t],t] 2000. {{s[t] -> -1600. + -------}} 0.05 t E so[t_] = s[t]/.sols[[1]]; And now we model the infielder's throw with initial velocity of 110 ft/sec and initial position of x feet from the plate. solv = DSolve[{v'[t] == - .05 v[t],v[0] == -110},v[t],t] -110 {{v[t] -> -------}} 0.05 t E vi[t_] = v[t]/.solv[[1]]; sols = DSolve[{s'[t] == vi[t],s[0] == x},s[t],t] 2200. {{s[t] -> -2200. + ------- + x}} 0.05 t E si[t_] = s[t]/.sols[[1]]; We determine the time it takes for the outfielder's throw to get to position x feet away from the plate (To(x)) and then we also determine the time it takes the infielder's throw to get from the position x feet away from the plate to the plate (Ti(x)). We then add these two times plus the 2 second time for the relay to take place to get the total time (T(x)) for the ball to go from outfielder at 400 feet to infielder at distance x to the plate. Further we presume the outfielder, infielder, and home plate are all in a line. solo = Solve[so[t] == x,t] Solve::ifun: Warning: Inverse functions are being used by Solve, so some solutions may not be found. {{t -> -20. Log[0.8 + 0.0005 x]}} To[x_] = t/.solo[[1]]; soli = Solve[si[t]==0,t] Solve::ifun: Warning: Inverse functions are being used by Solve, so some solutions may not be found. {{t -> -20. Log[1. - 0.000454545 x]}} Ti[x_] = t/.soli[[1]]; T[x_] = To[x] + Ti[x] + 2 2 - 20. Log[1. - 0.000454545 x] - 20. Log[0.8 + 0.0005 x] Notice that this equation makes sense with a reality check. For to have the outfielder throw 360 feet to the infielder when the infielder is 40 feet from the plate and the infielder throw the remainder is more time than to have the outfielder throw 40 feet to the infielder when the infielder is 360 feet from the plate and the infielder throw the remainder when the infielder can throw 110 ft/sec and the outfielder can only throw 100 ft/sec. I.e. we expect T[40] to be more than T[360] and it is. T[40] 6.336 T[360] 5.97789 (b)Find x which minimizes T. Plot[T[x],{x,0,400},PlotRange->{0,8}] Tder[x_] = T'[x] 0.00909091 0.01 ------------------ - -------------- 1. - 0.000454545 x 0.8 + 0.0005 x sol = FindRoot[Tder[x]==0,{x,210}] {x -> 300.} T[x]/.sol[[1]] 5.95794 Thus we see the best thing to do is to throw to the cutoff man at x = 250 feet from the plate and relay the ball from there for a total time of 6.20348 sec. (c) For a range of infielder capabilities, i.e., initial throwing velocities from 60 ft/sec to 140 ft/sec, prescribe the cutoff distance x and the total time for the relay to the plate. Now the infielder's throwing speed is a variable vio with range [60, 140] ft/sec. Thus we seek, T[x,vio] the total time with infielder throwing speed of vio at cut off distance. Thus we need to recompute the time it take the infielder to get the ball from distance x from the plate with throwing speed vio and add this time to the time the outfielder (outfielder's speed has not changed) takes to get the ball to distance x from the plate and the relay time of 2 seconds. solv = DSolve[{v'[t] == - .1 v[t],v[0] == -vio},v[t],t] vio {{v[t] -> -(------)}} 0.1 t E vi[t_] = v[t]/.solv[[1]]; sols = DSolve[{s'[t] == vi[t],s[0] == x},s[t],t] 10. vio {{s[t] -> -10. vio + ------- + x}} 0.1 t E sni[t_] = s[t]/.sols[[1]]; We solve for the time it takes to get to the plate, i.e. t such that sni[t] == 0. solni = Solve[sni[t]==0,t] Solve::ifun: Warning: Inverse functions are being used by Solve, so some solutions may not be found. 1. vio - 0.1 x {{t -> -10. Log[--------------]}} vio We now have the times for the new infielder speeds, vio. Tni[x_] = t/.solni[[1]]; And finally we have a new final time for the relay as a function of x (cutoff distance) and vio (infielder throwing speed). Tn[x_,vio_] = To[x] + Tni[x] + 2 1. vio - 0.1 x 2 - 10. Log[--------------] - 20. Log[0.8 + 0.0005 x] vio Now for each velocity vio we need to determine the cutoff distance, x, which yields the least time. Tnder[x_,vio_] = D[Tn[x,vio],x]; We define this function which gives the cutoff distance, x, and the total time of the relay throw as a function of the infielder's throwing speed, vio. p[vio_] := {x,Tn[x,vio]}/.FindRoot[Tnder[x,vio]==0, {x,100}] p[60] {-133.333, 6.19639} p[90] {66.6667, 6.41604} Recall our goal: For a range of infielder capabilities, i.e., initial throwing velocities from 60 ft/sec to 140 ft/sec, prescribe the cutoff distance x and the total time for the relay to the plate. Plot[p[vio][[2]],{vio,40,140},PlotRange->{0,8}, AxesLabel->{"speed","time"}] We plot the best cutoff distance x as a function of the infielder's throwing speed. From this we see that if the infielder throws at 80 ft/sec or less then the outfielder should throw the ball all the way home. And if the infielder can throw the ball at 140 ft/sec then the infielder should go out to meet the outfielder and throw the ball all the way in from 400 ft. Plot[p[vio][[1]],{vio,40,140}, AxesLabel->{"cutoff x","time"}] p[80] -7 {9.16124 10 , 6.46287} p[140] {400., 5.36472} (d) Consider the infielder who can throw 110 ft/sec. Would you rather improve the infielder's relay time, say down to 1.8 sec, or increase his initial velocity on the ball by 10 ft/sec? Explain. Tn[x,110] 110. - 0.1 x 2 - 10. Log[------------] - 20. Log[0.8 + 0.0005 x] 110 Plot[Tn[x,110],{x,0,400}] Tnder[x_] = D[Tn[x,110],x]; sol = FindRoot[Tnder[x]==0,{x,195,200}] {x -> 200.} Tn[x,110]/.sol[[1]] 6.11392 This is the minimum time if we can reduce the relay time by 0.2 sec. Tn[200,110]-.2 5.91392 We examine the case if we can increase the speed of our infielder's thrwo from 110 ft/sec to 120 ft/sec. Tnnder[x_] = D[Tn[x,120],x]; sol = FindRoot[Tnnder[x]==0,{x,195,200}] {x -> 266.667} Tn[x,120]/.sol[[1]] 5.893 Thus we see that we are better off working on increasing his speed than working on his relay time. ISSUES IN SOLUTION In part (c) we need to construct a complicated function which for each value of vio we chose we need to solve a transcendental equation to determine the cutoff distance x and hence the minimal time. This is not a closed form function and may be tricky for some students to work with, e.g. they may not be able to take the formal derivative.