Write/modify MATLAB code and create a properly labeled graph.Add 2 additional lines to your plot, “Cannon Range”, for v3 = 25 m/s and v4= 75 m/s BALLISTICSThe range of an object (see Figure 5.12) shot at an angle 0 with respect to the xaxisand an initial velocity th is given byR(0) =sin (20)for 0 s0s(neglecting air resistance)Use g= 9.9 m/s and an initial velocity of 100 m/s. Show that the maximumrange is obtained at 0 = 7/4 by computing and plotting the range for values ofO fromin increments of 0.05.Repeat your calculations with an initial velocity of 50 m/s, and plot both sets ofresults on a single graph.1. State the ProblemCalculate the range as a function of the launch angle.2. Describe the Input and OutputInputg= 9.9 m/s0 = 0 to 7/2, incremented by 0.0550 m/s and 100 m/s%3D Figure 5.12The ronge is zero, i the conon is perfecty vertical or perfecty horizontal.OutputRange RPresent the results as a plot.3. Develop a Hand ExampleIf the cannon is pointed straight up, we know that the range is zero, and if thecannon is horizontal, the range is also zero (see Figure 5.12).This means that the range must increase with the cannon angle up to somemaximum and then decrease. A sample calculation at 45 (/4 radians) showsthat- sin (20)%3D100sin9.9(等)= 1010 m when the initial velocity is 100 m/s

% Defining the paramtersv = 20.0;g = 9.8; % Create vector thetatheta = linspace(0,pi./2,63);R = zeros(63,1); % Create the output function for n = 1 : length(theta) a = (v.*v)./g; b = sin(2.*theta(n)); R(n) = a.*b;end % Printing the range Rdisp('R :')disp(R) %Plotting the relationfigure plot(theta,R);title('R is a function of theta');xlabel('theta');ylabel('R');hold on