Question

A skydiver of mass 77.0 kg jumps from a slow-moving aircraft and reaches a terminal speed of 53.8 m/s.

(a) What is her acceleration when her speed is 30.0 m/s?

magnitude	m/s2
direction	---Direction--- upward downward

(b) What is the drag force on the skydiver when her speed is 53.8 m/s?

magnitude	N
direction	---Direction--- upward downward

(c) What is the drag force on the skydiver when her speed is 30.0 m/s?

magnitude	N
direction	---Direction--- upward downward

Answer 1

ma = mg – ½ ρv²A(Cd)

where the drag force: Fd = ½ ρv²A(Cd)

a) First we need to find the surface area and drag coefficient product: A*Cd

At terminal speed, v' = 53.8 m/s, the drag force equals the skydivers' weight.

mg = Fd = ½ ρv²A(Cd)

77*9.81 = 0.5*1.29*53.8²[A*(Cd)]

A*(Cd) = 77*9.81/(0.5*1.29*53.8²) = 0.404

Now that we know A*(Cd) = 0.404 we can find the drag force at v = 30 m/s

Fd = ½ ρv²A(Cd) = 0.5*1.29*30²*0.404 = 234.52 N

The acceleration can be found using:

ma = mg - Fd

a = g - (Fd)/m = 9.81 - 234.52/77 = 9.81 - 3.04 = 6.76 m²/s , downward

b)  Fd = ½ ρv²A(Cd) = 0.5*1.29*53.8²*0.404 = 754.23 N, upwards

c) Fd = 0.5*1.29*30²*0.404 = 234.52 N, upwards