1. • The free falling velocity of a falling object is derived from Newton’s seco
ID: 1596808 • Letter: 1
Question
1. •
The free falling velocity of a falling object is derived from Newton’s
second law as:
v = gm(1 e(c/m)t) c
where v is the velocity, g is the gravitational force, m mass of the falling object, t time and c is the drag coefficient for this particular ob ject.
Lets assume we are trying to design a parachute that needs to free fall at a speed of 40 m/s after free falling for 10 seconds. The weight of the parachute user together with the parachute is calculated to be 68.1 kg. The acceleration due to gravity is taken as 9.81 m/s2. What should we take as the drag coefficient c (and hence determining the parachute) to attain this speed requirement?
a) write a matlab function that uses mynewton() as it is given in the book with 1,2,..,n iterations to find and return estimates c1,c2,...,cn as an array named c and the residual errors r1,r2,...,rn as an array named r. (Read myrootfind in the 18th page of the book to see how you can form these arrays).
b) prepare a table in a word processor (word, excel, libre office, open office, latex. you may also want to use export matlab arrays to excel) that showcases these values in a nice and coherent way for n=50.
c),d) repeat part a and part b with mybisect() as it is defined in the book. Also in part c) make your function calculate the error margins e1,e2,...,en and in part d) include error margin in the table (error margin of mybisect as it is defined in the book).
e) compare tables obtained in b) and d) write a few sentences to de- scribe the difference between Newton’s method and bisection method.
Send all relevant files via ”learn”, comment all of your code.
HINT: Graph the function whose root you are trying to find using matlab
to find a guess for the initial estimates.
HINT: If you copy paste mynewton and mybisect from the book you may have to fix a few errors with single quotation marks and comment marks in description of inputs .
Explanation / Answer
putting the values, we will get
0.0599*c+exp(-0.1468*c)-1=0
Solve this equation in matlab using function and iterative methods
take help of this program given below
function [ a,b ] = myrootfind(f,a0,b0 )
% function [a ,b ] = myrootfind (f ,a0 , b0 )
% Looks for subintervals where the function changes sign
% Inputs : f -- a function
% a0 -- the left edge of the domain
% b0 -- the right edge of the domain
% Outputs : a -- an array , giving the left edges of subintervals
% on which f changes sign
% b -- an array , giving the right edges of the subintervals
n = 50; % number of test points to use
a = []; % start empty array
b = [];
% split the interval into n -1 intervals and evaluate at the break points
x= linspace (0 ,1 ,n );
y=f(x);
% loop through the intervals
for i = 1:( n -1)
if y( i )* y(i +1) < 0 % The sign changed , record it
a = [a x( i )];
b = [b x( i +1)];
end
end
if size (a,1) == 0
warning (' no roots were found ')
end
try
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