A sky diver of mass 60 kg can slowherself to a constant speed of 93 km/hby orien

Question

A sky diver of mass 60 kg can slowherself to a constant speed of 93 km/hby orienting her body horizontally, looking straight down with armsand legs extended. In this position, she presents the maximumcross-sectional area and thus maximizes the air-drag force on her. (a) What is the magnitude of the drag force onthe sky diver?
1 N

(b) If the drag force is equal to bv2, what isthe value of b?
2 kg/m

(c) At some instant she quickly flips into a "knife" position,orienting her body vertically with her arms straight down. Supposethis reduces the value of b to 52 percent of the value in Parts (a) and (b). Whatis her acceleration at the instant she achieves the "knife"position?
3 m/s2 (a) What is the magnitude of the drag force onthe sky diver?
1 N

(b) If the drag force is equal to bv2, what isthe value of b?
2 kg/m

(c) At some instant she quickly flips into a "knife" position,orienting her body vertically with her arms straight down. Supposethis reduces the value of b to 52 percent of the value in Parts (a) and (b). Whatis her acceleration at the instant she achieves the "knife"position?
3 m/s2

Explanation / Answer

given mass of the diver is m = 60 kg the speed of the diver is V = 93 km /h = 25.833m/s as she is travelling with uniform velocity the net forceon her is zero i.e the weight is equal to the drag force                    F = mg = 60 kg * 9.8 m/s 2 = 588 N b ) given F = bv2 so           b = F / V2   = 0.88 c ) 52 % of b value will be = 0.52*0.88 = 0.45 so the drag force is F = 0.45*V2 =305.68 N so the net force is = 588 N - 305.68 N = 282.32N so the acceleration is = 282.32 N / 60 kg =4.705 m/s2 so           b = F / V2   = 0.88 c ) 52 % of b value will be = 0.52*0.88 = 0.45 so the drag force is F = 0.45*V2 =305.68 N so the net force is = 588 N - 305.68 N = 282.32N so the acceleration is = 282.32 N / 60 kg =4.705 m/s2

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