Plotting

How to plot trajectory of an object when the angle of launch varies from 15-75 using 15 degree increments. I have the code to plot it. However, all the 4 plots are blue in color. How do I make it display different colored plots for different angles?

Answer 1

[Walter Roberson]

plot(t,h15,'r');
hold on
plot(t,h30,'g');
plot(t,h45,'b');
plot(t,h60,'c');
plot(t,h75,'m');

Or

plot(t,h15,'r',t,h30,'g',t,h45,'b',t,h60,'c',t,h75,'m')

[Nishant Nain]

here's the code I have-

y0= 0; % m/s

v0=28; % m/s

g=9.8; % acceleration due to gravity in m/s^2

x=0:5:80; %distance from 0-100 in increments of 5

for th= 15:15:75;

y= (tand(th) .*x ) - (g.*x)/((2.*v0.^2).*(cosd(th)).^2) + y0;

plot (x,y);

legend (x,y);

xlabel('distance 0-80');

ylabel('vertical displacement y');

axis square;

hold on;

end

What should I change in this to make the plots look different.

[Nishant Nain]

^^ gives the plot with different angles but I can't get it to show different colors/forms for seperate values of th

Answer 2

[Oleg Komarov]

Vectorized version, you can see here that the color is automatically selected for each column of y

y0 = 0;
v0 = 28; 
g  = 9.8; 
x  = 0:5:80; 
th = 15:15:75;
y  = bsxfun(@times,tand(th), x.') - bsxfun(@rdivide, g*x.',2*v0^2*cosd(th).^2) + y0;
plot (repmat(x.',1,5),y);
xlabel('distance 0-80');
ylabel('vertical displacement y');
axis square

Note that you legend accepts strings not doubles.

Otherwise use the syntax that Walter suggested:

y0  = 0; % m/s
v0  = 28; % m/s
g   = 9.8; % acceleration due to gravity in m/s^2
x   = 0:5:80; %distance from 0-100 in increments of 5
th  = 15:15:75; 
clr = {'r','g','b','c','m'}; % Colors

h = axes;
axis(h,'square');
hold on
for t = 1:numel(th)
    y = tand(th(t))*x - g*x/(2*v0^2*cosd(th(t))^2) + y0;
    plot(x,y,clr{t});
end
xlabel('distance 0-80');
ylabel('vertical displacement y');

Some suggestions, don't abuse the parenthesis and try to understand when you really need .* instead of *.