Terminal velocity is the constant velocity an object falling through a fluid reaches when the upward drag force grows large enough to exactly balance gravity, making the net force and acceleration zero. On AP Physics C, you find it by setting F_drag = mg and solving for v.
Terminal velocity is what happens when drag wins the tug-of-war it was always going to win. When an object falls through air, gravity pulls down with a constant force mg, but the drag force pushes up and grows as the object speeds up. Eventually the drag force gets big enough to equal mg. At that moment the net force is zero, the acceleration is zero, and the object stops speeding up. That final constant speed is the terminal velocity.
In AP Physics C: Mechanics, terminal velocity is less about the definition and more about the math. You'll model drag with an expression like F_drag = bv or F_drag = bv², write Newton's second law as a differential equation (m dv/dt = mg − bv, for example), and find terminal velocity by setting the acceleration to zero. The velocity-vs-time graph is the signature picture. It starts with slope g (pure free fall for an instant), curves over, and approaches the terminal velocity as a horizontal asymptote without ever technically crossing it.
Terminal velocity sits with Topic 7.1, Gravitational Forces, because it's the realistic version of falling. Gravity provides the constant mg pulling down, and terminal velocity is what that fall looks like once you stop pretending air doesn't exist. It's also one of the best places in the course where calculus actually earns its keep. Newton's second law becomes a separable differential equation, and terminal velocity is the steady-state solution. If you can set net force to zero, solve for v_terminal, and sketch the exponential approach to it, you've demonstrated exactly the modeling skill Physics C is built around.
Keep studying AP Physics C: Mechanics Unit 7
Drag Force (Topic 7.1)
Drag is the velocity-dependent force that makes terminal velocity possible in the first place. Because drag grows with speed, a falling object always speeds up toward the point where drag catches gravity. Terminal velocity is just the speed where that catch-up happens.
Free Fall (Topic 7.1)
Free fall is the drag-free idealization where acceleration stays at g forever. Terminal velocity is the correction to that story. At the instant of release the object really is in free fall (v = 0, so drag = 0), which is why every velocity-time graph for a dropped object starts with slope g.
Air Resistance (Topic 7.1)
Air resistance is the specific fluid drag in most AP problems. Models like F = bv or F = bv² give different terminal velocities (mg/b versus √(mg/b)), and the exam expects you to handle whichever model the problem hands you rather than memorizing one formula.
Newton's Second Law as a Differential Equation
Terminal velocity is the classic Physics C differential equation setup. Writing m dv/dt = mg − bv, separating variables, and integrating gives v(t) as an exponential approach to v_terminal. This is the same math pattern you see in other resistive-force and damping problems across the course.
Terminal velocity shows up as a derivation task, not a vocab check. On the 2024 FRQ (Question 2), a cylinder is dropped from rest and the air exerts a drag force modeled as F_drag = b times a power of velocity. You're expected to draw a free-body diagram, write Newton's second law, set acceleration to zero to derive an expression for terminal velocity, and often sketch or interpret the v(t) graph that flattens toward it. Multiple-choice questions love the conceptual trap version. They'll ask for the acceleration at terminal velocity (it's zero), the acceleration at the instant of release (it's g, because v = 0 means no drag yet), or how terminal velocity compares for objects of different mass or drag coefficient. The move that earns points every time is setting the net force equal to zero and solving for v.
Free fall means gravity is the only force acting, so the acceleration is constant at g the whole way down. An object at terminal velocity is not in free fall at all. Drag has completely canceled gravity, so the acceleration is zero and the velocity is constant. Quick check for any falling-object problem: in free fall, a = g and v keeps growing; at terminal velocity, a = 0 and v stops changing. A dropped object actually transitions from one to the other, starting in essentially free fall and asymptotically approaching terminal velocity.
Terminal velocity is reached when the drag force grows to exactly balance the gravitational force, making the net force and acceleration both zero.
To find terminal velocity, set the magnitudes equal (F_drag = mg) and solve for v; with linear drag F = bv you get v_terminal = mg/b.
At the instant an object is dropped, v = 0 means drag is zero, so the initial acceleration is exactly g even when air resistance exists.
The velocity-time graph for a dropped object with drag starts with slope g, curves over, and approaches terminal velocity as a horizontal asymptote.
An object never technically reaches terminal velocity in the model; it approaches it exponentially, which is why the differential equation solution matters.
Heavier objects with the same drag coefficient have a larger terminal velocity, because it takes more drag (and therefore more speed) to balance a bigger mg.
Terminal velocity is the constant velocity a falling object reaches when the upward drag force equals the downward gravitational force mg. With zero net force, the acceleration is zero, so the object falls at a steady speed from then on.
Yes. At terminal velocity, drag exactly balances gravity, so the net force is zero and Newton's second law gives a = 0. That's the defining condition you use to solve for v_terminal on the exam.
Free fall means gravity is the only force, so acceleration is constant at g. Terminal velocity is the opposite extreme, where drag has fully canceled gravity and acceleration is zero. A real dropped object starts near free fall and asymptotically approaches terminal velocity.
Set the drag force magnitude equal to mg and solve for v. For F_drag = bv, terminal velocity is mg/b; for F_drag = bv², it's √(mg/b). The 2024 FRQ Question 2 used exactly this kind of setup with a dropped cylinder.
Not in the mathematical model. Solving m dv/dt = mg − bv gives a velocity that approaches v_terminal exponentially but never quite gets there. On graphs, draw terminal velocity as a horizontal asymptote the curve approaches, not a line it touches.
Connect this key term to the AP exam workflow: review the course, practice questions, and check related study tools.
Review units, study guides, and course resources.
Check this vocabulary in multiple-choice context.
Apply key concepts in written AP responses.
Estimate the exam score you are working toward.
Review the highest-yield facts before practice.
Put the full course together before test day.