How to calculate the jumper's final distance on this problem?

This is a Matlab Grader problem for calculating the jumper's final distance with given conditions. I wrote the code down below from my lecture and I got an answer of 7.3323, but it's not correct, can anyone take a look please? Thank you.

function Dist = BJump
    z0 = [0;0;pi/8;10];
    dt = 0.1;
    
    T = zeros(1,100);
    T(1) = 0;
    Z = zeros(4,100);
    Z(:,1) = z0;
    j = 1;
    
    %for j = 1:100-1
    while Z(2,j) >= 0
        K1 = physics(T(j),Z(:,j));
        K2 = physics(T(j) + dt/2,Z(:,j) + dt/2*K1);
        K3 = physics(T(j) + dt/2,Z(:,j) + dt/2*K2);
        K4 = physics(T(j) + dt,Z(:,j) + dt*K3);
        Z(:,j+1) = Z(:,j) + dt/6*(K1 + 2*K2 + 2*K3 + K4);
        T(j+1) = T(j) + dt;
        j = j + 1;
    end
    plot(Z(1,1:j),Z(2,1:j))
    x = Z(1,1:j);
    Dist = x(end)
    
    function dzdt=physics(t,z)
        dzdt = 0*z;
        dzdt(1) = z(4)*cos(z(3));
        dzdt(2) = z(4)*sin(z(3));
        dzdt(3) = -9.81/z(4)*cos(z(3));
        D = (0.72)*(0.94)*(0.5)/2*(dzdt(1)^2 + dzdt(2)^2);
        dzdt(4) = -D/80-9.81*sin(z(3));
    end
end

[Image Analyst]

Can you give us the correct answer?

[Zifeng Qiu]

I am sorry, I don't know what the correct answer is.

Answer 1

[David Hill]

function dist=BJump(v,theta,rho,s)
    x=0;
    g=9.81;
    c=.72;
    dt=.000001;%not sure how accurate you need
    y=v*sin(theta)*dt;
    while y>1e-7
        dTheta=-g*cos(theta)*dt/v;
        dv=(c*rho*s/2-g*sin(theta))*dt;
        x=v*cos(theta)*dt+x;
        v=v+dv;
        theta=theta+dTheta;
        y=y+v*sin(theta);
    end
    dist=x;
end

When I run this:

d=BJump(10,pi/8,.94,.5);%I get d=7.2748

[Zifeng Qiu]

Thank you