The data statistics in MATLAB plot does not show the correct value

Intro to engineering MATLAB courseware course.

I have a plot of a projectile, where y-axis is the height of the projectile, and x-axis is the time.

When I view the 'Data Statistics' in the drop down tools on the plot, it shows me the correct maximum value for y, but for the maximum value of x it just shows 20 for all the plots. 20 is the maximum the axis goes, but I need the time it took for the projectile to hit the ground, so the maximum x value for each plot. As you can see in the plot below for theta = 10, x should be less than 5 but it shows 20.

The code is:

% initial speed in m/s
V0 = 90;
% angle in degrees
theta = [10,25,45,65,85]';
% initial height above the ground in meters
y0 = 0;
% time in seconds
t = 0:0.01:20;
% gravity in m/s^2
g = 9.81;
% x position of the projectile in meters
xposition = (V0.*cosd(theta)).*t;
% y position of the projectile in meters
yposition = (-0.5*g.*(t.^2))+((V0.*sind(theta)).*t)+y0;
%% plotting the yposition of the projectile vs the time
plot(t , yposition)
xlabel('Time (s)')
ylabel('Projectile Height (m)')
ylim([0 500])
legend('Theta = 10', 'Theta = 25', 'Theta = 45', 'Theta = 65', 'Theta = 85')

Please could you help.

[Benjamin Thompson]

What is the data for X and Y? We cannot judge what the correct statistics are without the original data.

[Adam Danz]

My naive guess is that each of those lines continue to x=20 either at y=0 or below y=0 in which case, the max x value of 20 is correct. If this is the case, you can find the index value idx where the blue curve reaches y<=0, ignoring the initial coniditions, and then use that index value to get the time x(idx).

[Ghazwan]

can you share the code?

[dpb]

"My naive guess is that each of those lines continue to x=20 either at y=0 or below y=0 ..."

Nothing naive about it at all, Adam, the image above shows the min y for theta=10 is -1649.

OP forgot to stop the simulation for each angle at the time the projectile returned to earth and log the time that occurred. Or, as you note, he/she can now search/interpolate for the crossing point after the fact.

[Farooq]

I have added my code to the question.

[Farooq]

Could you please tell me how to stop this simulation when the projectile hits the ground, and the other method of indexing. I know how to use the min and max function when indexing but how do I get the index of the x value when y = 0. It will be two values because y = 0 at the x = 0, and y = 0 at x = 'projectile hits the ground'.

[Adam Danz]

Once the ball hits the ground, it's y value should remain at y=0 for the remainder of the time (assuming no bounce). So you'll see flat lines at y=0 after the curves reach the ground.

That's an easy change.

Hint:

max(yposition,0)

[Torsten]

Could you please tell me how to stop this simulation when the projectile hits the ground

Solve the equation

(-0.5*g.*(t.^2))+((V0.*sind(theta)).*t)+y0 = 0

for the nontrivial value of t (quadratic equation, should be no problem).