I am trying to find the roots of this equation but my code doesn't work. It keeps saying theta0 is unrecognized,

fun = @f;

x0 = [0 6.28];

z = fzero(fun,x0)

function theta0 = f(y0,yf,v0,x)

y0= 2;

yf= 1;

v0=20;

x=35;

theta0= x*tan(theta0)-4.905*(x^2)/((v0^2)*((cos(theta0))^2))-(y0-yf);

end

Answer 1

[Arif Hoq]

save your function. you have to define your variable theta0 in this function.

function y = f(x)
y0= 2;
yf= 1;
v0=20;
theta=35;
y= x*tan(theta0)-4.905*(x^2)/((v0^2)*((cos(theta0))^2))-(y0-yf);
end

then call this function in a diffrent script(m file)

fun = @f; 
x= [0 6.28]; 
z = fzero(fun,x)

[Anita Osoh]

x is a value in the function. I am solving for theta. That is why it has no variable.

Answer 2

[John D'Errico]

It appears that your equation is:

theta0= x*tan(theta0)-4.905*(x^2)/((v0^2)*((cos(theta0))^2))-(y0-yf);

and you wish to solve for x that satisfies this implicit function of x, where theta0 = 35 is given also?

The trick is to write it as a problem where the function would be zero. Do that by subtracting theta0 from both sides. Now your problem looks like:

x*tan(theta0)-4.905*(x^2)/((v0^2)*((cos(theta0))^2))-(y0-yf) - theta0 == 0

Now it is time to write some MATLAB code. We should verify it has any solution at all.

y0 = 2;
yf = 1;
v0 = 20;
theta0 = 35;
F = @(x) x*tan(theta0)-4.905*(x.^2)/((v0^2)*((cos(theta0))^2))-(y0-yf) - theta0;

PLOT IT!!!!

fplot(F,[-100,100])

So your problem, if it is assumed to be a function of x, has NO point where it crosses zero.

As such fzero would fail, although solve would find there are a pair of complex roots, since this is just a quadratic equation. THus

syms x
vpasolve(x*tan(theta0)-4.905*(x.^2)/((v0^2)*((cos(theta0))^2))-(y0-yf) - theta0)

ans = 

[Anita Osoh]

x is a value in the function. I am solving for theta. That is why it has no variable.

Answer 3

[Torsten]

function main
  theta0 = 0; 
  theta = fzero(@f,theta0)
end
function res = f(theta)
  y0 = 2;
  yf = 1;
  v0 = 20;
  x = 35;
  res = x*tan(theta)-4.905*(x^2)/((v0^2)*((cos(theta))^2))-(y0-yf);
end

[Walter Roberson]

I have a suspicion that they intend x to be a parameter, and need to know about parameterizing functions