Could anyone tell me why this keeps getting stuck in the first 'if' statement?

function [outcome,val] = Q2_13009636(v,theta,R,H)
    %gravity
    g = -9.81;
    outcome=4;

      while outcome==4;
      s = (R*tan(theta))* - ((g*R^2)/(2*v^2*(cos(theta))^2));

      if s>H; 
          outcome=2;
          val=s-H;
      end

          if H>=s; 
              if s>0;
                  outcome=1;
                  val=s;
              end
          end

                  if s<=0;
                      outcome=0;
                      val=0;
                  end
      end
  end %function end

Answer 1

[Star Strider]

Are ‘s’ or ‘H’ a vector?

Vector logic usually require the any or all functions to provide ‘correct’ comparisons.

[rlcal94]

So s is just the hight of the wall (H) plus the hight at which the object passes over it(h).

[Walter Roberson]

Please give us some sample inputs. Please also indicate more specifically what you mean when you say it is getting "stuck"

[rlcal94]

The only sample I have been given is 'v=28m/s, theta=pi/4 rad, R=75m, H=10m' and i've been told that this must give my second case 'if' statement of'outcome=1, val=s'

What I mean by getting stuck is that it drops into the first 'if' statment and outputs the 'outcome=2' and on top of that its not outputing a value for 'val'

[Star Strider]

The problem is that s=5278.8 which is greater than H=10. Your ‘s’ calculation may be in error.

[rlcal94]

You sir are a lifesaver, 100% on the dot. I bracketed it out wrong. Cannot believe I missed that.

[Star Strider]

My pleasure.

Believe me, it happens to us all! (Well, at least it’s happened to me.)

Answer 2

[Walter Roberson]

Your function uses "* - " in succession, so you are multiplying by the negative of something. That something has only positive components except for g, which is negative. negative times negative gives a positive, so the overall result will be positive unless tan(theta) is negative. And the value will probably be fairly large.

It appears to me that you wanted to multiply by something and subtract something else where the something else is expected to be positive.