function [root, fx, ea, iter] -bisect (func, xl, xu, es, maxit, varargin) bisect: root location zeroes % [root, fx, ea, iter]=bisect (func, xl, xu, es, maxit, p1, p2,...): % uses bisection method to find the root of func % input: % func = name of function % x1, xu = lower and upper guesses % es = desired relative error (default = 0.0001%) % maxit = maximum allowable iterations (default = 50) p1, p2,... = additional parameters used by func % output: % root = real root % fx = function value at root % ea = approximate relative error (%) %iter = number of iterations if nargin<3, error('at least 3 input arguments required'), end test = func (x1, varargin{:}) *func (xu, varargin{:}); if test>0, error('no sign change'), end if nargin<4|isempty(es), es=0.0001; end if nargin<5|isempty (maxit), maxit=50; end iter = 0; xrxl; ea = 100; while (1) end xrold = xr; xr = (x1 + xu)/2; iter = iter + 1; if xr = 0,ea = abs ((xrxrold)/xr) * 100; end test = func(x1, varargin{:}) *func (xr, varargin{:}); if test < 0 xu = xr; elseif test > 0 else end x1 = xr; ea = 0; if ea <=es | iter >= maxit, break, end root = xr; fx = func(xr, varargin{:}); Problem 05.001 - Solving a root problem using bisection method Use bisection method to determine the drag coefficient needed so that a 95-kg bungee jumper has a velocity of 46 m/s after 9 s of free fall. Star with initial guesses of x = 0.2 and x = 0.5 and iterate until the approximate relative error falls below 5% Note: The acceleration of gravity is 9.81 m/s². (Round the final answer to four decimal places.) The required drag coefficient is cd 0.4062 ± 0.0001 kg/m. Explanation: The function to evaluate is gm f (ca) =V tanh | v(t) m On substituting the given values f(ca)=√ 981(35) tanh 9.81c49-46 95 The first iteration is 0.2+0.5 x = 2 = 0.35 f(0.2)f (0.35) 12.706474 (2.3387193) = 29.716876 Therefore, the root is in the second interval and the lower guess is redefined as x, = 0.35. The second iteration is x = 0.35+0.5 2 = 0.425 0.425-0.35 E₁ = 0.425 x 100% 17.65% f(0.35)f (0.425) = 2.3387193 (-1.2809449) = -2.99577 i 1(x₁) Xu f(x) xx 1(x1) |εal f(x)(x) 1 0.2 12.706474 0.5 2 0.35 2.338719 0.5 3 0.35 2.338719 0.425 4 0.3875 0.434088 0.425 -4.2485678 -4.2485678 -1.2809449 -1.2809449 0.35 0.425 0.3875 0.40625 2.3387193 -1.2809449 0.4340883 -0.4452446 29.716876 17.65% 9.68% -2.995770 1.015211 4.62% -0.193275 Thus, after four iterations, a root estimate of 0.40625 is obtained with an approximate error of 4.62%, which is below the desired estimate of 5%.