Question

PART C PLEASE!!! IF YOU KNOW HOW TO CODE MATLAB PLEASE HELP THANK YOU

Write a Matlab/Python script which computes the trajectory of P according to the system of equations (1) and the corresponding (non-dimensional) range for given values of (α,φ,0), varying values of φ 0 Plot the trajectories for 6 values of φ 0 of your choice in the interval 0< φ0 π/2.

Compute the corresponding range reached by each trajectory.

Note: Special attention will be given to the computation of R within a specified tolerance.

b. It is convenient to use the following non- dimensional variables Show that the non-dimensional ODEs are given by dV do dT Vcos dY dT with the initial conditions V(0)-1, φ(0)-d, x(0) = Y(0) = 0 and parameter α-kt립mg Hence, in non-dimensional form, the study involves only 2 parameters: α and φο, we will fix the value of α to α-1.8, which corresponds to a baseball with speed 100 mph as it leaves the bat c. Write a Matlab/Python script which com putes the trajectory of P according to the sys tem of equations (1) and the corresponding (non- dimensional) range for given values of (a,Po), vary ing values of a-Plot the trajectories for 6 values of a) of your choice in the interval 0

Answer 1

Function file to solve ODE

funNondimenTraj.m

function fval= funNondimenTraj(T,K)

%% function file for ODEs i.e system of governing eqautions

alpha=1.8;

V=K(1);

phi=K(2);

X=K(3);

Y=K(4);

fval(1,1)=-sin(phi)-alpha*(V^2);

fval(2,1)=-cos(phi)/V;

fval(3,1)=V*cos(phi);

fval(4,1)=V*sin(phi);

end

Script file for solving and plotting Non dimensional number as $\Phi 0$ varies

runNondimenTraj.m

clc; clear all;

% Six values of Phi0

phi0=[pi/18, pi/9, pi/6, 2*pi/9, 5*pi/18, pi/3];

% plotting as phi0 varies in different subplot

for i=1:6

K0(:,i)=[1; phi0(i); 0; 0];

Tspan=[0 10];

[T,Ksol]=ode45(@(T,K) funNondimenTraj(T,K), Tspan, K0(:,i));

subplot(2,3,i), plot(T,Ksol);

legend({'V','phi','X','Y'});

xlabel('T');ylabel('Non dimensional Parameters');

hold on;

end

shg

% displaying min and max values of V, Phi, X, Y at end value of i

% so change i in for loop for differnet values of phi0

V=Ksol(:,1);

Vmin=min(V);

disp(['min value of V =',num2str(Vmin)]);

Vmax=max(V);

disp(['max value of V =',num2str(Vmax)]);

phi=Ksol(:,2);

phiMin=min(phi);

disp(['min value of phi =',num2str(phiMin)]);

phiMax=max(phi);

disp(['max value of phi =',num2str(phiMax)]);

X=Ksol(:,3);

Xmin=min(X);

disp(['min value of X =',num2str(Xmin)]);

Xmax=max(X);

disp(['max value of X =',num2str(Xmax)]);

Y=Ksol(:,4);

Ymin=min(Y);

disp(['min value of Y =',num2str(Ymin)]);

Ymax=max(Y);

disp(['max value of Y =',num2str(Ymax)]);

% subplot titles

subplot(2,3,1),title('pi/18');

subplot(2,3,2),title('pi/9');

subplot(2,3,3),title('pi/6');

subplot(2,3,4),title('2*pi/9');

subplot(2,3,5),title('5*pi/18');

subplot(2,3,6),title('pi/3');

Graphs

Non dimensional Parameters Non dimensional Parameters Non dimensional Parameters Non dimensional Parameters Non dimensional Parameters Non dimensional Parameters 8 2

Ranges of Non dimensional numbers When T varies from 0 to 10 at 6 different values of $\Phi 0$

$\Phi 0$	$\pi /18$	$\Phi 0$	$\pi /9$	$\Phi 0$	$\pi /6$	$\Phi 0$	$2\pi/9$	$\Phi 0$	57/18	$\Phi 0$	π/3
V	0.5826 to 1	V	0.5349 to 1	V	0.4812 to 1	V	0.4221 to 1	V	0.3538 to 1	V	0.3538 to 1
Phi	-1.5708 to 0.1745	Phi	-1.5708 to 0.3491	Phi	-1.5708 to 0.5236	Phi	-1.5708 to 0.6981	Phi	-1.5708 to 0.8727	Phi	-1.5708 to 0.8727
X	0 to 0.7906	X	0 to 0.7857	X	0 to 0.7578	X	0 to 0.7050	X	0 to 0.6254	X	0 to 0.6254
Y	-6.8760 to 0.0126	Y	-6.7665 to 0.0434	Y	-6.6585 to 0.0844	Y	-6.5558 to 0.1303	Y	-6.4619 to 0.1775	Y	-6.4619 to 0.1775