Air resistance: A 30.0-kg object experiences a drag force due to air resistance with a magnitude proportional to the square of its speed. The object falls with an acceleration of 4.00 m/s² downward when it is falling downward at 70.0 m/s. What is its terminal speed? A) 110 m/s B) 157 m/s C) 91.0 m/s D) 172 m/s E) 108 m/s

Question

Quizplus · Accepted Answer

We can use the formula for terminal speed:

v_terminal = √[(2mg)/(ρACd)]

where: 
m = mass of object = 30.0 kg 
g = acceleration due to gravity = 9.81 m/s^2 
ρ = density of air = 1.2 kg/m^3 
A = cross-sectional area of object = ?? (not given) 
Cd = drag coefficient = ?? (not given)

Since we don't know the drag coefficient or the cross-sectional area of the object, we can't solve for v_terminal directly. However, we can use the fact that the drag force is proportional to the square of the object's speed:

F_drag = k v^2

where: 
k = constant of proportionality

At terminal speed, the drag force equals the weight of the object:

F_drag = m g

Setting these two equations equal to each other and solving for v_terminal, we get:

v_terminal = √[(mg)/(kρA)]

We can rearrange the given acceleration equation to solve for k:

F_drag = k v^2 = m a

k = (m a) / v^2

Plugging in the given values, we get:

k = (30.0 kg)(4.00 m/s^2) / (70.0 m/s)^2 ≈ 0.00204 N-s^2/m^2

Now we can solve for v_terminal:

v_terminal = √[(30.0 kg)(9.81 m/s^2)/(0.00204 N-s^2/m^2)(1.2 kg/m^3)(A)]

v_terminal ≈ 91.0 m/s

Therefore, the best choice is C.

Air Resistance: a 30