Question

The person’s mass, m= 68.1 kg, the unknown drag coefficient is, c , with units kg/s, and the local acceleration of gravity is g= 9.80 m/s2 . Our model for the person’s velocity vs time gave us a linear differential equation whose analytical solution was the following:
V(t) = g*m/c*[1-exp(-c/m*t)].

Your job is to find “c”. You need to use the MATLAB “help” documentation to find the nonlinear curve fitting function and syntax for how to use it. Hint you should use the documentation and example for “lsqcurvefit”.
Include in your program a plot of the experimental velocity data vs time and the calculated velocity vs time. The graph should have a title, x and y labeled axis, and a legend. It is likely your legend will not be placed in a good position. Use again the MATLAB “help” documentation to learn how to move it to the southeast corner inside your plot.

The following table is experimental data regarding our parachutist. We have a measured velocity versus time after the jump, but before the chute is opened:
Time t (sec)

Velocity, cm/s
0


1
1000
2

1630

3
2300
4

2750
5

3100

6
3560
7

3900

8
4150
9

4290

10
  
4500

11
4600
12

4550

13
4600
14

4900

15
5000

Answer 1

clc
clear all

m = 68.1;
g = 9.81;

tdata = 0:15;
vdata = [0 1000 1630 2300 2750 3100 3560 3900 4150 4290 4500 ...
4600 4550 4600 4900 5000 ];

plot(tdata, vdata,'or','MarkerFaceColor','g');
fun =@(c,tdata) 100*(g*m/c)*(1- exp(-c*tdata/m)); % 100 is multiplied to
% convert m/s to cm/s

c = lsqcurvefit(fun,20,tdata,vdata);
fprintf('c = %0.5f\n',c);

t = linspace(0,15);
v = fun(c,t);
hold on
plot(t,v,'-b','LineWidth',2);
xlabel('times( sec )');
ylabel('v(t) cm/s');
title('Parachute velocity vs time');
legend('Experimental curve','Fitted curve','Location','southeast');