Couple projectile motion equations function

I am creating a function to solve these coupled equations, where all constants are known:

I am currently using the code below, but I am not getting the correct results. All of the constants are specified in my code, b If anyone could offer any help that would be great!

% Set constants for different projectiles (baseball, football, soccer ball)

global m cd A rho g theta v0 x0 y0

m= [0.145 0.42 0.43];    % mass of projectile [kg]
cd= [0.3 0.5 0.25];    % drag coefficients 
A= [0.00426 0.02318 0.038]; % cross sectional area of projectile [m^2]
V= [0.0002194 0.00462115 0.00573547];   % volume of projectile
rho= [m(1)/V(1) m(2)/V(2) m(3)/V(3)];   % density 
g= 9.8; % gravity [m/s^2]
theta= 45;  % initial launch angle [rad]
x0= [0 0 0];    % initial horizontal position
y0= [1.8288 1.8288 0];  % initial vertical position
v0= [18 18 22]; % initial velocity (m/s) 

%set projectile, 1=baseball, 2=football, 3=soccer ball
proj=2;
    m= m(proj);
    cd= cd(proj);
    A= A(proj);
    rho= rho(proj);
    x0= x0(proj);
    y0= y0(proj);
    v0= v0(proj);

% call function
t_end= 2*v0*sin(theta)/g;   % time until projectile reaches the ground

[t y]= ode45(@my_fun,[0 t_end],[x0 y0 v0 theta]);


% plotting results
figure(1)
plot(t,y(:,1))
grid on
xlabel('time [s]')
ylabel('horizontal position [m]')
title('plot of horizontal location vs time')

figure(2)
plot(t,y(:,2))
grid on
xlabel('time [s]')
ylabel('vertical position [m]')
title('plot of vertical location vs time')

 figure(3)
 plot(y(:,1),y(:,2))
 grid on
 xlabel('horizontal position [m]')
 ylabel('vertical position [m]')
 title('plot of vertical location vs horizontal position')
 
%% Function
function dy=my_fun(t,y) % y is a matrix containing [x y v theta]
global m cd A rho g theta x0 y0 v0
dy=zeros(4,1);
dy(1)= y(3);   %dx/dt
dy(2)= y(3);   %dy/dt
dy(3)= -(cd*A*rho)/(2*m)*sqrt((dy(1))^2+(dy(2))^2)*dy(1);   % d2x/dt2
dy(4)= -g-(cd*A*rho)/(2*m)*sqrt((dy(1))^2+(dy(2))^2)*dy(2);  % d2y/dt2
end

Answer 1

[Ameer Hamza]

In your my_fun you have a line

dy(2)= y(3);   %dy/dt

It should be

dy(2)= y(4);   %dy/dt

Everything else is correct according to the equation in your question.

[Nicholas Moose]

It looks like y(4) is a launch angle. Shouldn't it be something like dy(1) = y(3)*cos(y(4)) and dy(2) = y(3)*sin(y(4))?

[James Tursa]

No. There are four states: x, y, vx, vy. These are defined to be y(1), y(2), y(3), and y(4) in the code. The dx/dt value is always going to be vx, which is simply y(3), and the dy/dt value is always going to be vy, which is simply y(4).

The first dx/dt and dy/dt values may in fact correspond to a launch angle, but that is only for the initial point.

That being said, the initial conditions that are passed to ode45( ) do not involve theta directly as was written. They should only be [x0 y0 vx0 vy0] where all of these values are scalars. So in this case since we are working with launch angle theta it would be:

[x0 y0 v0*cosd(theta) v0*sind(theta)]

Also this

theta= 45;  % initial launch angle [rad]

should be this

theta= 45;  % initial launch angle [deg]

and downstream in the code you should be using sind(theta) instead of sin(theta).