Can someone help with my code I know MATLAB well I want the code to run and print out a time and height graph.

h

[Image Analyst]

Sorry, my MATLAB doesn't do OCR to turn your image into code. I can't run the image directly. So, you'll have to post the code in your message or attach the m-file with the paper clip icon. In the meantime, look at the fprintf() function.

[Honore Lopaka]

could not find a way to send the file i have attached a word document

[Mathieu NOE]

hello

would be nice if you could precise which variables you want to print out

also I have a doubt that the code is valid , how is t1 calculated ?

[Honore Lopaka]

[Honore Lopaka]

i want to do that in matlab i can solve it by hand i just dont know how to solve it using matlab and then print out h and t graph

[Image Analyst]

Attach your m-file with the paper clip icon. Or else just copy from MATLAB and paste back here in your reply, then highlight it and click the code icon.

Answer 1

[Mathieu NOE]

so -- after several corrections, we hav something

see below

Also I noticed in the hand paper you have swapped the initial values for V1x and V1y

clc;
close all;
clear all;

%User Input
h_person_in = (51.3/12);% HEIGHT in inches times .75 then converted to feet
v1 = 10; %ft/s current velocity
theta_deg = 15; %degrees
e = .8; %coefficient of restitution
x = 6.10 % ft length of where i want the ball to land
g = 32.2 % ft/s^2
t =0 % starting time (initial conditions)

% v1y = v1*sin(theta_deg);
% v1x = v1*cos(theta_deg);
v1y = v1*sin(theta_deg*pi/180); %correction 
v1x = v1*cos(theta_deg*pi/180); %correction 

% v1y^2 == v1y^2+(2*-g*h_person_in)
v1y_squared = v1y^2+(2*g*h_person_in);  % correction 
v1y = sqrt(v1y_squared); % missing line

v2 = sqrt((e*v1y)^2+v1x);

% solve for t1 (second order equation)
% h_person_in = v1y*t1+(.5*g*t1^2)
% reorgainzed like : ax² + bx + c = 0
a = .5*g;
b = v1y;
c = - h_person_in;

% determinant : delta = b² -4*a*c
delta = b^2 -4*a*c;
% solution (positive) : sol = (-b+sqrt(delta)) / (2*a)
t1 = (-b+sqrt(delta)) / (2*a);


theta_deg2 = 180/pi*atan((e*v1y)/v1x);  % correction 

x1= v1x*t1;
x2= x-x1;       
% tf= v1x/x2;  % time cannot be velocity / distance it's the contrary
tf= x2/v1x;    % 


% display (example) in command window
disp([' time t1 is : ' num2str(t1) ' seconds']);
disp([' time tf is : ' num2str(tf) ' seconds']);