Chair Acceleration Part 2?

Question

A sky diver of mass 80.0 kg (including parachute) jumps off a plane and begins her descent.
Throughout this problem use 9.80 m/s^2 for the magnitude of the acceleration due to gravity.

For an object falling through air at a high speed , the drag force acting on it due to air resistance can be expressed as

F = Kv^2

where the coefficient K depends on the shape and size of the falling object and on the density of air. For a human body, the numerical value for K is about 0.250 kg/m.

Using this value for K, what is the terminal speed v-terminal of the sky diver?
Express you answer in meters per second.

B) When the parachute is fully open, the effective drag coefficient of the sky diver plus parachute increases to 60.0 kg/m. What is the drag force F(drag) acting on the sky diver immediately after she has opened the parachute?

C) What is the terminal speed v(terminal) of the sky diver when the parachute is opened?
Express your answer in meters per second.

Answer 1

The terminal velocity is the velocity when the falling object stops accelerating (i.e. the velocity stops changing)

F = MA so A = F/M

If the object's velocity isn't changing, the acceleration A = 0 so the net force F also has to be 0.

The net force downward F = W - D
where W is the weight (i.e. the force due to gravity), and
D is the drag force. F = 0 when D = W

We know the mass of the sky diver and so can compute W, the force due to gravity.

Since we know K, we can compute the V^2 that makes D = W, and hence V, the desired terminal velocity.

With that V and the new K (the drag coefficient for the open parachute), we can compute the new drag force.

Eventually, the parachute will slow the diver to a new, slower terminal velocity, but we still have the new K so we can compute it.