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WileyPLUS M rirefox Browser Privacy Noi X | + https://edugen.Wileyplus.com/edugen/student/mainfr.uni C aSearch Boyco/DiPrima/Meade, Elementary Differential Equations with Boundary Value Problems, 11e DIFFERENTIAL EQUATIONS (MATH 2120/4391) Home Read, Study & Practice Assignment> Open Assignment ] ASSIGNMENT RESOURCES Assignment Gradebook Downloadable eTextbook FRSTON BACK NEXT Chapter 2, Section 2.3, Additional Question 04 Section 2.3 Chapter 2, Sertinn 23, A skydiver weighing 299 lb (including equipment) falls vertically downward from an altitude of 4000 ft and opens the parachute after 13 s of free fall. Assume that the force of air resistance, which is directed opposite to the velocity, is 0.74|v1 when the parachute is closed and 101v1 when the parachute is open, where the velocity v is measured in ft/s. E Chaprer2, Section 2..3, Use g = 32 ft/s?, Round your answers to two decimal places. (a) Flnd the speed of the skydlver when the parachute opens. r(13) = (b) Fird Ihet dislance allen before lhe# perachuLe uper is. Chapter 2, Saction 2.3, Additional Question 01 hapter 2, Section2 dditional Ou x(13) ft Review Results by Study Objective (c) What Is the lImiting veloclty VL after the parachute opens? ft/s Click if ynl. would liktc) Sh w work fr this quAstinn: By accessing this Question Assistance, you will learn while you earn points based on the Point Potential Policy set by your instructor 9/13/2017

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Dear Student Thank you for using Chegg !! Given a skydriver with a parachute a) Velocity when parachute opens (After 13 sec) V = u + at u = -0.74 ft / sec a = g (32 ft/s^2) t = 13 s (given) v= 13 * 32 416 ft/sec b) Distance travelled before parachute opens s = ut + (1/2)at^2 u '=0 t = 13 sec a = g (32 ft / sec2) s = (1/2)(32)(13)^2 = 2704 ft c) Limiting velocity After parachute opens, distance travelled 1296 ft v^2 - u^2 = 2as u '= 416 ft / sec s = 1296 ft a = g v^2 = 256000 v = 505.9644256 ft/sec Hence limiting velocity is 505.9 ft/ sec

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