Problem 5 (computer) An aerodynamic equation for free fall that includes the effects of air resistance tells us that tha

[answerhappygod]

Problem 5 Computer An Aerodynamic Equation For Free Fall That Includes The Effects Of Air Resistance Tells Us That Tha 1 (36.11 KiB) Viewed 50 times

Problem 5 Computer An Aerodynamic Equation For Free Fall That Includes The Effects Of Air Resistance Tells Us That Tha 2 (58.18 KiB) Viewed 50 times

Problem 5 (computer) An aerodynamic equation for free fall that includes the effects of air resistance tells us that that in a time 1 seconds a certain object falls a distance y meters where y=-Incosh (t√gk)] and g=9.81 m/s.k=0.00341 m. Write a program named FullName5.sce that uses the secant method to determine the amount of time required to fall 1 kilometer. Find with enough precision that the corresponding value of y equals 1 km to within 1 cm. Your program should print the value of t to the console. Problem 6 (computer) The vibration frequencies of a beam satisfy the equation tan(o)tanh(0)+1=0 Write a program named FullName6.sce that starts at co=2 and iterates Newton's method until a solution is found with a precision of better than six decimal places. Your program should print the value of to the console.
9.1 Newton's method $rootNewton.m 2014-06-04, Scott Hudson $ Implements Newton's method for finding a root f(x) = 0. Requires two functions: y=f(x) and y=fp (x) where fp (x) is the derivative of f(x). Search starts at x0. Root is returned as r. $888 function r = rootNewton (x0, f, fp, tol) MAX_ITERS = 40; give up after this many iterations nIters = 1; 1st iteration. r = x0-f(x0) /fp (x0); %%Newton's formula for next root estimate while (abs (r-x0) >tol) & (nIters<-MAX_ITERS) nIters nIters+1; keep track of # of iterations x0= r; current root estimate is last output of formula r = x0-f(x0) / fp (x0); Newton's formula for next root estimate end end 9.2 Secant method 8888 rootSecant.m ** 2014-06-04, Scott Hudson Implements secant method for finding a root f(x) = 0. * Requires two initial x values: x1 and x2. Root is returned as r accurate to (hopefully) about tol. function r root Secant (x1, x2, f, tol) MAX_ITERS = 40; nIters = 1; fx2= f(x2); r = x2-fx2* (x2-x1) / (fx2-f(x1)); while (abs (r-x2) >tol) & (nIters<-MAX_ITERS) nIters+1; nIters x1 = x2; end &maximum number of iterations allowed 1st iteration fx1 = fx2; x2 = x; fx2= f(x2); r = x2-fx2* (x2-x1) / (fx2-fx1); end