Gravity, drag and terminal velocity
You will be able to: Solve a constant driving force opposed by linear resistance and interpret terminal motion.
How can a falling object reach a steady speed?
A falling object speeds up until resistance nearly balances its weight. At terminal velocity it is still moving, but its acceleration approaches zero.
A useful starting point: Solve a velocity-dependent resistive force →
Words and symbols before equations
- Downward-positive velocity v
- Positive v represents falling in this lesson.
- Terminal velocity v_t
- Steady velocity where the driving force and resistance balance.
- Asymptote
- A limiting value approached by the model.
- τ = m/b
- Time constant for the linear-drag model.
What this picture assumes
Release from rest, downward positive, gravity plus linear drag, g = 10 m/s². Terminal velocity is approached asymptotically. No collision surface is modeled.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- v_t = 5 m/s down; τ = 0.5 s. At 0.5 s: v = 3.161 m/s, a = 3.679 m/s² down, displacement 0.9197 m down.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
Take down positive: m dv/dt = mg − bv. Balance gives v_t = mg/b. Let u = v_t − v. Then du/dt = −(b/m)u, which has the exponential form from the previous lesson.
For release from rest, v = v_t[1 − e^(−t/τ)] and a = ge^(−t/τ). Integrating from y(0) = 0 gives downward displacement y = v_t[t − τ(1 − e^(−t/τ))]. At t = 0 these satisfy zero speed and acceleration g.
More generally v = v_t + (v₀ − v_t)e^(−t/τ). If released faster than v_t, the object slows toward it; terminal speed is not a universal upper bound for arbitrary initial conditions. At terminal motion the forces are nonzero and balanced. A different drag law changes the approach and terminal value.
A worked example, step by step
A 2 kg object is released from rest with linear resistance b = 4 kg/s. Find terminal speed, time constant and speed at 0.5 s using g = 10 m/s².
- v_t = mg/b = 20/4 = 5 m/s downward.
- τ = m/b = 0.5 s.
- At t = τ, v = 5(1 − e⁻¹) ≈ 3.161 m/s.
- Acceleration is 10/e ≈ 3.679 m/s² downward; it has not reached the limiting zero acceleration.
Terminal velocity means zero net force, not zero gravitational force or zero velocity.
If initial speed is above terminal speed, does the object keep accelerating down?
Compare with an explanation
No. Drag exceeds weight, producing upward acceleration until speed approaches terminal from above.
Predict. Change one thing. Explain.
Increase time and compare weight with drag. Then increase b at fixed mass. Predict terminal speed and the time constant separately.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
v_t = 5 m/s down; τ = 0.5 s. At 0.5 s: v = 3.161 m/s, a = 3.679 m/s² down, displacement 0.9197 m down.
Release from rest, downward positive, gravity plus linear drag, g = 10 m/s². Terminal velocity is approached asymptotically. No collision surface is modeled.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant force, system boundary, acceleration or calculus relationship to justify your prediction.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionA 1 kg object falls from rest with b = 2 kg/s. Find v_t, τ, v(1 s) and acceleration at 1 s.
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Compare with the answer and four-point rubric
- 1 point: v_t = 10/2 = 5 m/s down.
- 1 point: τ = 1/2 = 0.5 s.
- 1 point: v(1) = 5(1 − e⁻²) ≈ 4.323 m/s.
- 1 point: a(1) = 10e⁻² ≈ 1.353 m/s² down.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1Why does acceleration decrease during a fall from rest?
Drag increases as speed grows, reducing the net downward force.
RECALL 2What balances at terminal velocity?
The constant driving force and resistance.
RECALL 3Can an object start faster than terminal speed?
Yes. It then approaches terminal speed from above in this model.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Gravity, drag and terminal velocity
- Down positive: m dv/dt = mg − bv.
- v_t = mg/b; v = v_t(1 − e^(−t/τ)) from rest.
- y = v_t[t − τ(1 − e^(−t/τ))]; τ = m/b.
Remember: Terminal velocity means zero net force, not zero gravitational force or zero velocity.
Conditions: Release from rest, downward positive, gravity plus linear drag, g = 10 m/s². Terminal velocity is approached asymptotically. No collision surface is modeled.
Refresh Kid · AP Physics C: Mechanics Unit 2 (official Unit 2) · Objectives 2.9.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 2.9, objectives 2.9.A. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026 alongside the Fall 2026 clarifications. This is Mechanics Unit 2: Force and Translational Dynamics. The unit covers Topics 2.1–2.10. Calculus is introduced where it is needed for continuous mass and velocity-dependent forces. Shell theorem is applied without requiring a proof; spring combinations are purely series or purely parallel. The lesson breakdown and questions are original Refresh Kid work, not official topic subdivisions.
Implementation and automated checks are separate from independent teacher review and observation of students. Both human review stages remain pending. This is a review edition, not a certified or validated assessment.
Optional further resource: College Board’s released questions and scoring guides. Papers can combine units; this link is an archive, not an assignment of every question to this lesson.
Our learn, explore, practice and recall sequence is informed by the IES learning guide. The exact Refresh Kid implementation has not been evaluated for learning effectiveness.
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