Topping the Green Monster Matlab Help

Topping the Green Monster

The Green Monster is a wall 37 ft high in left field at Fenway Park in Boston. The wall 310 ft from home plate down the left-field line, Assuming that the batter hits the ball above the ground, and neglecting air resistance, determine the minimum speed the must give to the ball to hit it over the Green Monster. In addition, find the angle at ~
the ball must be hit (see Figure 10.3-1) .

• Solution
The equations of motion for a projectile launched with a speed Vo at an angle () relati the horizontal are
x(t) = (vocos(  gt2 y(t) = — + (vo sin (})t


where x = 0, y = 0 is the location of the ball when it is hit. Because we are not concerned with the time of flight in this problem, we can eliminate t and obtain an equation for y in terms of x. To do so, solve the x equation for t and substitute this into the y equation to obtain g x2(t) y(t)=– + x(t)tan B 2 V6 cos? B (You could use MATLAB to do this algebra if you wish. We will use MATLAB to do the more difficult task to follow.) Because the ball is hit4ft above the ground, the ball must rise 37-4 = 33 ft tocIearthe wall. Let h represent the relative height of the wall (33 ft). Let d represent the distance to the wall (310 ft), Use g = 32.2 ft/sec”, When x = d, y = h. Thus the previous equation gives g d2 h = — +dtanB 2 v6 cos? B which can easily be solved for V6 as follows: 2 g d2 Vo =”2 cos2B(dtanB – h)
Because Vo > 0, minimizing V6 is equivalent to minimizing Vo. Note also that gd2/2 is a multiplicative factor in the expression for V6′ Thus the minimizing value of B is independent of g and can be found by minimizing the function

f I =  cos? B(d Ian B – h)

The session to do thi is as follows. The variable th represents the angle 0 of the balJ’s velocity vector relative to the horizontal. ‘The first step is to calculate the derivative df IdO and solve the equation dfidO = 0 for O.
»syrns d g h th »f = 1/(((cos(th»A2)*(d*tan(th)-h»; »dfdth diff(f,th); »thmin solve(dfdth,tb); »thmin subs (thmin, {d.h},{310,33}) thmin = 0.8384 -0.7324
Obviously, the negative angle is not a proper solution, so the only solution candidate is 0 = 0.8384 rad, or about  8°. To verify that this angle is a minimum solution, and nol a maximum or an inflection point, we can check the second derivative d2 f Id02. If thi derivative is positive, the solution represents a minimum. To check this solution and to find the speed required, continue the session as follows: »second diff(f,2,·thL % This is the second derivative. »second = subs(second,{th,d,h},{thmin(1),310,33}) second = 0.0321
»v2 = (g*dA2/2)*f; »v2rnin = subs(v2,{d,h,g},{310,33,32.2}); »vrnin = sqrt(v2rnin); »vrnin = double (subs (vrnin(l),{th,d,h,g}, {thmin(l) ,310,33,32.2}) vrnin = 105.3613
Because the second derivative is positive, the solution is.a minimum. Thus the mini- JDJml speed reqWred is 1053613 filsec, or.about 72 miIhJ:. A hall hit”.Jbi.s speed dear lhe wall only if it is hit at .aD angle of approximatciy 48


Posted on July 30, 2015 in Symbolic Processing with MATLAB

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