We were asked to plot the trajectory of a diver off a diving board, and I am having trouble getting my graph to publish for this problem. Can anyone tell me what I have done wrong? Thank you.
if true
t=2.5; %in seconds
Vox=2
Voy=8.25
tx=0:.1:2.5; %this will give me an array for the time of the diver in the x position
ty=0:.1:2.5; %this will give the array for the time of the diver in the y-position
function [tx,ty]= trajectory(Vox,Voy,tx,g)%function for varying angle
x=Vox*cos(0)*tx; %gives the position of the x
y=Voy*(sin(0)*ty)-(.5*g*(ty.^2)); %gives the position of the y
plot (x,y)
end
end

Réponses (1)

Star Strider
Star Strider le 4 Oct 2018

0 votes

Your ‘trajectory’ function wants ‘ty’ as well, so you need to include that in the argument list. Other than that, you need to call it correctly.
Try this:
g = 9.81;
t=2.5; %in seconds
Vox=2;
Voy=8.25;
tx=0:.1:2.5; %this will give me an array for the time of the diver in the x position
ty=0:.1:2.5; %this will give the array for the time of the diver in the y-position
function [tx,ty]= trajectory(Vox,Voy,tx,ty,g)%function for varying angle
x=Vox*cos(0)*tx; %gives the position of the x
y=Voy*(sin(0)*ty)-(.5*g*(ty.^2)); %gives the position of the y
plot (x,y)
end
[tx,ty] = trajectory(Vox,Voy,tx,ty,g);
When I run it, it produces an acceptable plot.
Also, if you want ‘trajectory’ to return the calculated values for ‘x’ and ‘y’, you need to tell it.
Consider:
function [x,y] = trajectory(Vox,Voy,tx,ty,g)%function for varying angle

1 commentaire

Drishya Dinesh
Drishya Dinesh le 2 Déc 2020
Modifié(e) : Drishya Dinesh le 2 Déc 2020
can i get a time varying trajectory for an aerial vehicle in 3d (6 dof motion- x,y,z,phi,theta,psi)? Even helical is fine

Connectez-vous pour commenter.

Catégories

En savoir plus sur Networks dans Centre d'aide et File Exchange

Tags

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by