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

At terminal speed, the net force on the object is zero (since it is no longer accelerating).
The drag force due to air resistance is proportional to v^2, so we can write:
F_drag = kv^2
where k is a constant of proportionality.
At terminal speed, the drag force is equal in magnitude but opposite in direction to the weight of the object:
F_drag = F_weight
kv^2 = mg
where g is the acceleration due to gravity.
Solving for v, we get:
v = sqrt(mg/k)
Substituting in the given values, we get:
v = sqrt(30.0 kg * 9.81 m/s^2 / (4.00 m/s^2 * 70.0 m/s))
v ≈ 91.0 m/s

A 300-Kg Object Experiences a Drag Force Due to Air Resistance