finding intercept point in the plot

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theta = 50;
v0=200;
slope= 0.1;
vx=v0*cosd(theta); %horizontal component of velocity vector
vy=v0*sind(theta); %vertical component of velocity vector
g=9.807;
tf=2*vy/g; %total flight time
t=linspace(0,tf,150);
for r=1:length(t)
x(r)=vx*t(r); %horizontal postion
A(r)=vy*t(r)-0.5*g*t(r)^2; %vertical postion of particle
R(r)=x(r)*slope; %vertical postion of terrain
end
figure(1)
plot(x,A,'b')
hold on
plot(x,R,'m')
title('Altitude VS Range')
xlabel('Range')
ylabel('Altitude')
legend({'Range','Line of Terrain'},'Location','southwest')
grid on
hold on

Accepted Answer

Star Strider
Star Strider on 8 Apr 2021
Try this:
theta = 50;
v0=200;
slope= 0.1;
vx=v0*cosd(theta); %horizontal component of velocity vector
vy=v0*sind(theta); %vertical component of velocity vector
g=9.807;
tf=2*vy/g; %total flight time
t=linspace(0,tf,150);
for r=1:length(t)
x(r)=vx*t(r); %horizontal postion
A(r)=vy*t(r)-0.5*g*t(r)^2; %vertical postion of particle
R(r)=x(r)*slope; %vertical postion of terrain
end
[Amax,idx] = max(A);
x_int = interp1(A(idx:end)-R(idx:end), x(idx:end), 0);
y_int = interp1(x, R, x_int);
figure(1)
plot(x,A,'b')
hold on
plot(x,R,'m')
plot(x_int, y_int, 'rs')
title('Altitude VS Range')
xlabel('Range')
ylabel('Altitude')
legend({'Range','Line of Terrain','Intercept'},'Location','southwest')
grid on
hold off
.
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More Answers (2)

FRANCISCO CORTEZ
FRANCISCO CORTEZ on 12 Apr 2021
How can i determine the initial angle of the projectile with respect to the horizontal in order to achieve maximum range

Image Analyst
Image Analyst on 22 May 2022
See attached demo that computes just about everything you could ever want to know about a projectile.

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