An object with mass of 90kg is projected upward form the surface of
the earth at a velocity of 60m/s. If the Earth is subject to linear
drag(c=15kg/s), use the golden section search to determine the
maximum height the object attains.
In addition to the numerical solution, determine the analytic
solution and compare the two values.
Determine the time of flight (use a root finding method)
Plot y vs t during the time of flight.
`Hey,
Note: 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
format long;
c=15;
v0=60;
m=90;
g=9.81;
f=@(t) -((m/c)*(v0+m*g/c)*(1-exp(-c*t/m))-m*g*t/c);
a=0; % start of interval
b=10; % end of interval
epsilon=0.000001; % accuracy value
iter= 50; % maximum number of iterations
tau=double((sqrt(5)-1)/2); % golden proportion coefficient, around
0.618
k=0; % number of iterations
x1=a+(1-tau)*(b-a); % computing x values
x2=a+tau*(b-a);
f_x1=f(x1); % computing values in x points
f_x2=f(x2);
while ((abs(b-a)>epsilon) && (k<iter))
k=k+1;
if(f_x1<f_x2)
b=x2;
x2=x1;
x1=a+(1-tau)*(b-a);
f_x1=f(x1);
f_x2=f(x2);
else
a=x1;
x1=x2;
x2=a+tau*(b-a);
f_x1=f(x1);
f_x2=f(x2);
end
k=k+1;
end
% chooses minimum point
if(f_x1<f_x2)
sprintf('t_max=%f', x1)
sprintf('f(t_max)=%f ', -f_x1)
else
sprintf('t_max=%f', x2)
sprintf('f(t_max)=%f ', -f_x2)
end
disp('Time of flight is');
t_flight=fzero(f,10)
fplot(@(t) -f(t),[0,t_flight]);
disp('Analytical solution is');
tm=fminbnd(f,0,10);
disp(-f(tm));

Kindly revert for any queries
Thanks.
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