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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

David Hill
on 12 Jul 2020 at 1:47

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

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