Question

A sky diver of mass 53 kg can slow herself to a constant speed of 92 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.

1) a) What is the magnitude of the drag force on the sky diver?

b) If the drag force is equal to bv², 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 50 percent of the value in Parts (a) and (b). What is her acceleration at the instant she achieves the "knife" position?

Answer 1

help from

A sky diver of mass 66 kg can slow herself to a constant speed of 99 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) Fairly simple problem here.   If the diver has constant speed then the drag force is equal and opposite to the force of gravity. Drag = m *g = 66kg * 9.8m/s^2 up = 647 n up

(b) Be careful regarding the units in part b. Note the speed is in km/hr.

bv^2 = mg;   So b = mg/v^2 =   9.8m/s^2 * 66kg/(99km/hr)^2 * (1km/1000m)^2 *(3600s/1hr)^2

b = 0.85 kg/m

(c) if b is reduced by 0.54 of original avlue then the drag force becomes 0.54 * 647n or 349n.    

The net force on the sky diver is then gravity force - drag force = 9.8m/s^2 * 66kg - 349n = 298 n downward.

The acceleration a = F/m 298n/66kg = 4.5m/s^2 downward.