Question

Assume that y' = y.t needs to solved i.e.one needs to find y(t) for the initial condition yo = 0.3. In your Math classes, you were taught how to solve this differential equation. However, most differential equations are not easily solvable, and one needs to resort to numerical integration. This can easily be done in MATLAB for explicit first order differential equations like the one stated above. The time of interest is between tstart <t <tend let's say between O and 2 seconds. Using the following MATLAB syntax, this will lead to a plot of the y(t) relationship: yo = 0.3; timeSpan = [02]; [Time, Yout] = ode45(@(t.y)(y*t),time Span,yo): plot(Time, Yout); Now, assume a parachutist with a mass m of 80kg, a cross sectional area of 0.65 m², a drag coefficient Cp = 0.45 at an atmospheric density at sea level p = 1.225kg/m has an initial velocity of O m/s after dropping out of a balloon. Numerically solve the equation below, plot the velocity-time relationship, and determine the terminal velocity. Also, discuss the formula with your peers. Can you find the terminal velocity analytically? mV' = m.g-V.S.Cd

Answer 1

`Hey,

Note: Brother if you have any queries related the answer please do comment. I would be very happy to resolve all your queries.

clc%clears screen
clear all%clears history
close all%closes all files
m=80;
S=0.65;
g=9.81;
Cd=0.45;
p=1.225;
f=@(t,v) (m*g-p/2*v^2*S*Cd)/m;
[T,V]=ode45(f,[0,25],0);
plot(T,V);
disp('Terminal velocity is');
disp(V(end));

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Kindly revert for any queries

Thanks.