I am wanting to delete these entries of vector x when x = 0. How do I do this. Matlab gives me an error right now saying index cannot exceed 38 but I don't know what's wrong.

clear all 
for i = 10:1:65
    
    
    x(i-9) = i;
    theta = i;
    tspan = linspace(0,1000,1000);
    V = 60;
    Vx = 60*cosd(theta);
    Vy = 60*sind(theta);
    options.Events = @event_func;
    [t,f,te,fe,ie] = ode45(@(t,f) eqs_of_motion(t,f),tspan,[0 0 Vx Vy],options);
    if fe(:,2) > 25
        x(i-9) = x(i-9);
        y(i-9) = fe(:,2);
    else fe(:,2) <=25
        x(i-9) = 0;
        y(i-9) = 0;
    end
end

Unrecognized function or variable 'eqs_of_motion'.

Error in solution>@(t,f)eqs_of_motion(t,f) (line 12)
[t,f,te,fe,ie] = ode45(@(t,f) eqs_of_motion(t,f),tspan,[0 0 Vx Vy],options);

Error in odearguments (line 92)
f0 = ode(t0,y0,args{:}); % ODE15I sets args{1} to yp0.

Error in ode45 (line 104)
odearguments(odeIsFuncHandle,odeTreatAsMFile, solver_name, ode, tspan, y0, options, varargin);

for i = 1:56
    if x(i) == 0
        x(i) = []
    end
end 

Answer 1

[Torsten]

Use

x(x==0) = [];

instead of

for i = 1:56
    if x(i) == 0
        x(i) = []
    end
end 

The length of the x-vector will become shorter than 56 in the course of the loop such that x(56), e.g., will no longer exist when i = 56. This causes an access error.

Answer 2

[Walter Roberson]

@Torsten's recommendation of x(x==0) = [] is the best. But alternately,

for i = 56:-1:1
    if x(i) == 0
        x(i) = []
    end
end 

Each time you delete an entry, the following entries "fall down" to fill the hole. If you are looping forward 1:56 then if even one entry gets deleted then x(56) will no longer exist and you get the error. If you loop backwards then you do not encounter the same problem.