A sky diver of mass 61 kg can slow herself to a constant speed of 95 km/h by ori

Question

A sky diver of mass 61 kg can slow herself to a constant speed of 95 km/h by orienting her body horizontally, looking straight down with arms and legs extended. In this position, she presents the maximum cross-sectional area and thus maximizes the air-drag force on her. (a) What is the magnitude of the drag force on the sky diver? N (b) If the drag force is equal to bv2, what is the value of b? kg/m (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 53 percent of the value in Parts (a) and (b). What is her acceleration at the instant she achieves the "knife" position? m/s^2

Explanation / Answer

speed is constant

aceeleration = a = 0

Fnet = Fd - Fg

Fnet = m*a

Fd - Fg = 0

Fd = Fg

Fd = mg

Fd = 61*9.8 = 597.8 N <------------answer

(b)

Fd = bv^2

v = 95 km/h = 95*(5/18) = 26.4 m/s

597.8 = b*26.4^2

b = 0.86 kg/m <<-------------answer

(c)

if b' = 0.53b = 0.53*0.86

Fnet = Fg - Fd = m*a

(61*9.8)-(0.53*0.86*26.4^2) = 61*a

acceleration a = 4.6 m/s^2 <<<---------answer

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