Adjust the launching angle to produce the greatest range (i.e. maximum horizontal distance) but within 0° ≤ 𝜃଴ ≤ 90° in incremental of 0.1. Evaluate your results.

I need help trying to code this,

The pdf attached shows the questions, question 2vi is a problem because when i plot it into a graph theres so many lines that its a blur to try and determine which angle creates the furthest horizontal distance. please someone help!!

[Steven Lord]

Since this is a class assignment, please show us the code you've written to try to solve the problem (the one that creates "so many lines that its a blur") and we may be able to provide some guidance.

[Kara Streader]

attached here

[Walter Roberson]

You plot() a 2d array. When you do that, one line is created for each column of the array. Try transposing the arrays

plot(x.', y.')

Answer 1

[David Hill]

%% Question v
Y = @(t,theta)y0 - 1/2*g*t.^2 + (v0*sind(theta))*t;
X = @(t,theta)x0 + (v0*cosd(theta))*t;
%% Question vi
theta=0:.1:90;
a = -1/2*g; 
b = v0*sind(theta);
c = y0;
ytminus = ((-b) - sqrt((b.^2)-4*a*c))/(2*a);
[~,idx] = max(x0 + (v0*cosd(theta)).*ytminus);
%%
figure(f);hold on;%previous f=figure; (for the figure to reuse)
plot(X(t,theta(idx)),Y(t,theta(idx)));

[David Hill]

@ is a function handle and in this case indicates an anonymous function. See: https://www.mathworks.com/help/matlab/ref/functions.html

or

https://www.mathworks.com/help/matlab/matlab_prog/anonymous-functions.html

[Kara Streader]

ah perfect thank you

[Kara Streader]

this doesn't seem to be working - is there any other way to do this?

[David Hill]

I cleaned up your code a little bit. This should work for you.

x0 = 0; 
y0 = 100;
t = 0;
g = 9.8;
v0 = 35;
Theta0 = 5*pi/12; 

%% Question i
y = @(t,theta)y0 - 1/2*g*t.^2 + (v0*sin(theta))*t;
x = @(t,theta)x0 + (v0*cos(theta))*t;
t = 0:.01:10;
figure;
plot(t,x(t,Theta0)); 
xlabel ('Time in seconds' )
ylabel ('Horizontal distance in metres')
grid on
figure;
plot(t,y(t,Theta0)); 
xlabel ('Time in seconds')
ylabel ('Vertical distance in metres')
grid on 
%% Question ii
a = -1/2*g; 
b = v0*sin(Theta0);
c = y0;
ytplus = ((-b) + sqrt((b^2)-4*a*c))/(2*a); 
ytminus = ((-b) - sqrt((b^2)-4*a*c))/(2*a);
xii = x0 + (v0*cos(Theta0))*ytminus;

%% Question iii
f=figure; 
plot(x(t,Theta0),y(t,Theta0));
grid on 
xlabel ('Horizontal Distance in Metres')
ylabel ('Vertical Distance in Metres')
%% Question iv
Theta1 = Theta0 - 0.255;
Theta2 = Theta0 - 0.425;
figure;
plot(x(t,Theta0),y(t,Theta0),x(t,Theta1),y(t,Theta1),x(t,Theta2),y(t,Theta2));
grid on
xlabel ('Horizontal distance in metres')
ylabel ('Vertical distance in metres')

%% Question v
Y = @(t,theta)y0 - 1/2*g*t.^2 + (v0*sind(theta))*t;
X = @(t,theta)x0 + (v0*cosd(theta))*t;
%% Question vi
theta=0:.1:90;
a = -1/2*g; 
b = v0*sind(theta);
c = y0;
ytminus = ((-b) - sqrt((b.^2)-4*a*c))/(2*a);
[~,idx] = max(x0 + (v0*cosd(theta)).*ytminus);
%%
figure(f);hold on;%previous f=figure; (for the figure to reuse)
plot(X(t,theta(idx)),Y(t,theta(idx)));

[Kara Streader]

How come we don't need for loops? That's all i've really been told (thank you so much by the way for helping out)

[Steven Lord]

Many functions in MATLAB can operate on arrays of data all at once and so don't need to be called once per element. For example if I wanted to take the sine of a vector of angles:

x = 0:0.1:2*pi;

I could do it one at a time:

y = zeros(size(x)); % Make an array of 0's the same size as x
for whichOne = 1:numel(x)
    y(whichOne) = sin(x(whichOne)); % Fill in that array of 0's
end

Or since the sin function is vectorized I could do it in one fell swoop.

z = sin(x);

The two approaches give the same answer.

isequal(y, z) % true

ans = logical
   1

[David Hill]

The power of matlab is in the use of matrx and array operations. If you examine the below code:

Y = @(t,theta)y0 - 1/2*g*t.^2 + (v0*sind(theta))*t;%anonymous function with inputs of time-array and theta
X = @(t,theta)x0 + (v0*cosd(theta))*t;%anonymous function
theta=0:.1:90;%array of theta [0 .1 .2 .3 .4 .... 90]
a = -1/2*g;%scalar
b = v0*sind(theta);%array since theta is an array sind(theta) produces an array
c = y0;%scalar
ytminus = ((-b) - sqrt((b.^2)-4*a*c))/(2*a);
% .^2 is an element-wise operation and squares each element of the b-array
% sqrt takes the sqrt of all elements of the array. -b-array - sqrt-array performs
% element-wise substraction (arrays must be the same size). /(2*a) divides
% each element by (2*a).
[~,idx] = max(x0 + (v0*cosd(theta)).*ytminus);% max finds the maximum value of the array (which we don't care about) and the index of the maximum value.
plot(X(t,theta(idx)),Y(t,theta(idx)));
%plots alls X values for the given time-array using the theta indexed at
%the maximum horizontal distance previously determined and similarily all
%corresponding Y values.

When using arrays, no for-loops are needed in the above code. If individual calculations were done, you would have to cycle through all the theta values going from 0 to 90 degrees.

Answer 2

[Kara Streader]

sorry i'm new to matlab, what are the @ symbols representing?

Answer 3

[Torsten]

To answer vi), use ii) to determine t when the ball hits the ground for a given value of theta. Then, for this value for t, determine x.

If you do this for the 901 theta values, you'll get a vector of 901 corresponding x-values.

Determine the maximum x value within this vector and the corresponding theta value (use max).

Plot the x-y-trajectory within the graph from iii) (use hold on/off)