A sky diver of mass 53 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 55 percent of the value in Parts (a) and (b). What is her acceleration at the instant she achieves the "knife" position?
m/s2

force = 520 N

b = 0.747 kg/m

acceleration = 4.41 m/s²    

Explanation:

given data

mass = 53 kg

speed = 95 km/h

solution

as here driver is at constant speed so here drag force is

drag force = mg

force = 53 × 9.81

force = 520 N

and when drag force is equal to bv²

b =  

b =  

b = 0.747 kg/m

and

here x = 520 × 0.55  = 286 N

so here

F net = 520 - 286

F net = 234 N

so here

acceleration =  

acceleration =  

acceleration = 4.41 m/s²    

if they have a finite resultant

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west