How to calculate the jumper's final distance on this problem?
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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
采纳的回答
David Hill
2020-7-12
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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