I have this code
clcclearclosea = 40*(pi/180);v = 1200;g = 32.2;k=[0,2e-6,10e-6,20e-6];k=0;dt=.5;t = 0:dt:55;v = zeros(length(t),4);vx = zeros(length(t),4);vy= zeros(length(t),4);x = zeros(length(t),4);y = zeros(length(t),4);v(1,1) = 1200;vx(1,1) = v(1,1)*cos(a);vy(1,1) = v(1,1)*sin(a);x(1,1)=0;y(1,1)=0;for i=2:length (t) x(i,1) = x(i-1,1)+vx(i-1,1)*dt-.5*k*v(i-1,1).^2*cos(a)*dt.^2; y(i,1) = y(i-1,1)+vy(i-1,1)*dt-.5*k*v(i-1,1).^2*sin(a)*dt^2-0.5*g*dt.^2; vx(i,1) = vx(i-1,1)-k*v(i-1,1).^2*cos(a)*dt; vy(i,1) = vy(i-1,1)-k*v(i-1,1).^2*sin(a)*dt-g*dt; a=atan(vy(i,1)./vx(i,1)); v(i,1)=sqrt(vx(i,1).^2+vy(i,1).^2); endvx(1:round(5/dt):end,:)vy(1:round(5/dt):end,:)x(1:round(5/dt):end,:)y(1:round(5/dt):end,:)
That results in
ans = 919.25 0 0 0 919.25 0 0 0 919.25 0 0 0 919.25 0 0 0 919.25 0 0 0 919.25 0 0 0 919.25 0 0 0 919.25 0 0 0 919.25 0 0 0 919.25 0 0 0 919.25 0 0 0 919.25 0 0 0ans = 771.35 0 0 0 610.35 0 0 0 449.35 0 0 0 288.35 0 0 0 127.35 0 0 0 -33.65 0 0 0 -194.65 0 0 0 -355.65 0 0 0 -516.65 0 0 0 -677.65 0 0 0 -838.65 0 0 0 -999.65 0 0 0ans = 0 0 0 0 4596.27 0 0 0 9192.53 0 0 0 13788.80 0 0 0 18385.07 0 0 0 22981.33 0 0 0 27577.60 0 0 0 32173.87 0 0 0 36770.13 0 0 0 41366.40 0 0 0 45962.67 0 0 0 50558.93 0 0 0ans = 0 0 0 0 3454.23 0 0 0 6103.45 0 0 0 7947.68 0 0 0 8986.90 0 0 0 9221.13 0 0 0 8650.35 0 0 0 7274.58 0 0 0 5093.81 0 0 0 2108.03 0 0 0 -1682.74 0 0 0 -6278.52 0 0 0
My issues is whenever I change the K value to [2e-6,10e-6, or 20e-6] it results in the first column changing to the respective k values. However, if I change the x(i,1) positions to say x(i,2) for [2e-6] (and all subsequent notations) then I get every k = 2e-6 value in the 2nd column and the first column becomes zero. Question is how do I get the answers to add on to the table made from previous calculations?
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