How to display values from a chart into the command window

I am modeling a 3d projectile. I was wondering how to get the end coordinates of the plot to appear in the command window so I dont have to manually find them. Here is the code. I want the values of x,y,z.

figure('name','Projectile Motion')
xlabel('X (m)')
ylabel('Y (m)')
zlabel('Z (m)')
% This sets the viewing angle
view([1 -1 1])
% This sets the limits on the x- and y-axes
xlim([0,3000])
ylim([0,3000])
zlim([0,1000])
box('on')
grid('on')
hold on
% az is azimuth angle, el is elevation
V0 = 150;
az = 45;
el= 30;
t = [0:0.01:25]';
v = sqrt(vx0^2+vy0^2+vz0^2)
vx0 = V0*cosd(az)*cosd(el);
vy0 = V0*sind(az)*cosd(el);
vz0 = V0*sind(el);
x = vx0.*t;
y = vy0.*t;
z = vz0*t-0.5*9.81.*t.^2;
plot3(x,y,z)
%Time of flight
T =(2*vz0)/9.81

[Daniel M]

You have the values already. Remove the semi colon to see them in the command window.

[Stephen Mixon]

I just want the final values after the projectile lands and z=o, when I remove the semicolon is gives me hundreds of values for x, y, and z.

Answer 1

[Daniel M]

If you want the end value, then do

coord = [x(end) y(end) z(end)]

If you want the value when z = 0, assuming it equals exactly zero,

ind = find(z==0,1,'last');
coord = [x(ind) y(ind) z(ind)]

If you want the closest data point to zero,

[~,ind] = min(abs(z));

If you want to choose the point graphically, select the data tip in the plot, right click and select export cursor to workspace.

[Stephen Mixon]

I tried all 3 and number 2 and 3 resulted in my coordinates coming out to all zeros. When I used the first one my z came out not at zero. How do I make sure the projectile stops at 0.

[Daniel M]

Method 1 didn't work because you computed your answer for too long and z was negative at the end. Method 2 didn't work because you're not computed z at a timepoint when it equals exactly zero (except at the beginning). And method 3 didn't work because z = 0 at the beginning.

[Daniel M]

There's a zillion ways to find x,y,z when z = 0 at the end of the trajectory.

If z(end) is positive, then increase t and keep computing, or use interp1 with the 'extrapolate' option.

If z(end) is negative, you can use

find(diff(sign(z)))

to find the locations of the zero crossings. From there you can also choose to interpolate the values to get the values when z=0.

Or you can use method 3 from above, but only after the trajectory has peaked.

[~,locmax] = max(z);
[closestToZero,locmin_tmp] = min(abs(z(locmax+1:end)));
locmin = locmax + locmin_tmp;

Or you can use your physics knowledge to solve for t when z = 0 and compute the solution at that time.