Question

A skydiver of mass 82.0 kg jumps from a slow-moving aircraft and reaches a terminal speed of 49.9 m/s.

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

(b) What is the drag force on the skydiver when her speed is 49.9 m/s?

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

Answer 1

Gravitational acceleration = g = 9.81 m/s²

Mass of the skydiver = m = 82 kg

Terminal speed of the skydiver = V_T = 49.9 m/s

Drag force is directly proportional to the square of velocity.

Drag force = F = kV² (k is the proportionality constant)

When the skydiver reaches terminal speed the drag force acting on the skydiver is equal to the weight of the skydiver.

Drag force acting on the skydiver at terminal speed = F₁

F₁ = mg

F₁ = (82)(9.81)

F₁ = 804.42 N

F₁ = kV_T²

804.42 = k(49.9)²

k = 0.323 N.s²/m²

Drag force on the skydiver when her speed is 30 m/s = F₂

Acceleration of the skydiver when her speed is 30 m/s = a₂

Speed of the skydiver = V₂ = 30 m/s

F₂ = kV₂²

F₂ = (0.323)(30)²

F₂ = 290.7 N

ma₂ = mg - F₂

(82)a₂ = (82)(9.81) - (290.7)

a₂ = 6.26 m/s²

a) Acceleration of the skydiver when her speed is 30 m/s = 6.26 m/s²

b) Drag force on the skydiver when her speed is 49.9 m/s = 804.42 N

c) Drag force on the skydiver when her speed is 30 m/s = 290.7 N