How Does Euler's Method Help Predict the Ball's Landing in Matlab Simulation?

[TriclesCr]

Homework Statement

Question :

The ball is kicked at a 5m high vertical wall 20m away. The initial speed of the ball is Vm/s and its initial angle of motion is C degrees with the horizontal (both the speed and angle are inputs). When the ball strikes the wall it bounces back with coefficient of restitution e , eventually returning to ground between the wall and the kicker.

The forces acting on the ball are gravity and a wind that acts at right angles to the initial motion of the ball, which generates a force of WN
The ball has diameter of 1 metre and weighs 0.5 kg.
Using Euler’s method,determine the location where the ball strikes the ground.

2. The attempt at a solution

I've never used MATLAB before,so i was looking for a tutorial about euler's method on MATLAB and i found this one : http://www.me.ucsb.edu/~moehlis/APC591/tutorials/tutorial1/node4.html

So i came up with this :

First i divided my work into 2 phases,first phase is when the ball hits the wall,the second one is when the ball bounces back and hit the ground.

%input
m = 0.5; %weighs
g = 9.81; % gravity
w = 1.06; %Wind Force
h = 0.5; %timestep
e = 0.78; % coefficient of restitution
npoints = 41;
%since my stationary point is at the centre of the ball so the distance between the Stationary point to the wall is now 20.5m(41xtimestep(0.5)=20.5)

V = input('Initial Speed ');
C = input('Initial Angle ');
V;
C;

%Equation for the first phase (when ball hits the wall)
% Centre of the ball as a stationary point
% y''(doubledot)=-g
% x''(doubledot)=0
% z''(doubledot)=w/m


%reduce to first oder equations
% let m = y' , s = x' , r = z'


x =zeros(npoints,1);%this initialize the vector y to being all zeros
y =zeros(npoints,1);
z =zeros(npoints,1);
m =zeros(npoints,1);
s =zeros(npoints,1);
r =zeros(npoints,1);

%Initial Condition
x(1) = 0;
y(1) = 0;
z(1) = 0;
m(1) = V*sind(C); %ydot(0)
s(1) = V*cosd(C); %xdot(0)
r(1) = 0; %zdot(0)




for n=1:npoints-1 %loop over the timesteps

m(n+1) = m(n) + h*-g;
y(n+1) = y(n) + h*m(n);

s(n+1) = s(n) + h*0;
X(n+1) = x(n) + h*S(n);

r(n+1) = r(n) + h*(w/m);
z(n+1) = z(n) + h*r(n);

end

plot(x,y,z);

==================================================
is that code right ?
I have not tested yet because i don't have MATLAB on my laptop,i did that one notepad.
If that is right,
How do i calculate the new velocity of the ball when it bounces back using e (coefficient of restitution ) ?
Once I get the new velocity,do I need to do the same step as above with different initial condition ?

Thanks

 

[KataKoniK]

in advance



Your code looks correct for the first phase of the ball hitting the wall. To calculate the new velocity after the bounce, you will need to use the equation for the coefficient of restitution, which is given by:

Vn = e*Vp

Where Vn is the new velocity and Vp is the previous velocity.

To incorporate this into your code, you will need to change the initial conditions for the second phase to be the final values from the first phase. So, for example, your new initial conditions for the second phase would be:

x(1) = x(npoints);
y(1) = y(npoints);
z(1) = z(npoints);
m(1) = -e*m(npoints); %use the new velocity for y
s(1) = s(npoints); %x velocity remains the same
r(1) = r(npoints); %z velocity remains the same

Then, you can use the same loop as before to calculate the trajectory of the ball for the second phase. Keep in mind that the ball will continue to bounce back and forth until it eventually comes to rest, so you may want to add a condition in your loop to stop the calculation once the ball reaches a certain height (e.g. less than 1 cm).

I hope this helps. Good luck with your calculations!