A skydiver of mass 60.0kg can slow herself to a constant speed of 90km/h by orie

Question

A skydiver of mass 60.0kg can slow herself to a constant speed of 90km/h by orienting her body horizontally, looking straight down with arms and legs extended. (a) What is the magnitude of the drag force on the skydiver? (b) If the drag force is given by bv^2, what is the value of b? (c) At some instant she quickly flips into a "knife" position, orienting her body vertically with her arms straight down. Suppose this reduces the value of b to 55 percent of the value in Parts (a) and (b). What is her acceleration at the instant she achieves the "knife" position? (d) With this new value of b what will her new terminal velocity be?

Explanation / Answer

a) Since the diver is falling at constant velocity drag force = gravitational pull drag force = mg =60*9.8 = 588N b)Drag force = bv^2 v= 90km/hr = 90*1000/3600 = 25m/sec drag force = b*25^2 = 588 => b = 0.94 c)b reduces to 55% new B = .94*.55 = .517 let acclimation at this instant be a now ma = mg - .55mg => a = .45g = 4.41 At the terminal velocity BV^2 = .55mg => V^2 = 60*9.8/.517 = 1137.33075 => V= 33.72m/sec 33.72*3600/1000 = 121.39 km/hr

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