Terminal Velocity (Air Resistance)
Key idea: Learn how air resistance affects a falling object and why terminal velocity happens when drag equals weight (O Level Physics 6091).
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The core idea
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Learning objectives
- Distinguish contact forces from non-contact forces
- State that mass measures the amount of matter in a body
- Describe a gravitational field as a region where a mass experiences gravitational force
- Define gravitational field strength as gravitational force per unit mass
- Apply weight = mass × gravitational field strength
- Distinguish mass from weight
- Describe the effect of balanced and unbalanced forces on a body
- Describe ways a force may change motion
- Identify action–reaction pairs on interacting bodies
- Draw free-body diagrams for force systems in at most two dimensions
- Solve three-force static equilibrium graphically
- Apply resultant force = mass × acceleration
- Relate mass to resistance to change in motion
- Explain the effects of friction on motion
- Describe falling with and without air resistance, including terminal velocity
- Describe a moment as a force's turning effect in everyday examples
- Apply moment = force × perpendicular distance from the pivot
- State the principle of moments for a body in equilibrium
- apply the principle of moments to new situations or to solve related problems
- show an understanding that the weight of a body may be taken as acting at a single point known as its centre of gravity
- Explain qualitatively how centre-of-gravity position affects stability
1. Definition
When an object falls through air, it experiences:
- weight, W = mg (downwards)
- air resistance (drag), F_d (upwards, opposite to the motion)
Terminal velocity is the constant velocity reached when the resultant force is zero:
F_d = W ⇒ a = 0
2. Key Ideas
- At the start: speed is low → drag is small → resultant force is downward → the object accelerates.
- As speed increases: drag increases → resultant force decreases → acceleration decreases.
- At terminal velocity: drag = weight → resultant force = 0 → constant velocity.
- A larger surface area (e.g. parachute) increases drag, giving a smaller terminal velocity.
- Terminal velocity depends on the object (shape, surface area, mass) and the fluid (air density) (qualitatively).
3. Detailed Explanations
You should be able to describe the motion of a falling object with or without air resistance, including terminal velocity.
A. Forces on a falling object
The two main forces are:
- weight W = mg (downwards, approximately constant near Earth)
- drag F_d (upwards, increases as speed increases)
The resultant force while falling is:
Taking downwards as positive:
Fᵣₑₛᵤₗₜₐₙₜ = W - F_d
You can choose any positive direction, but be consistent.
A common choice is downwards as positive, so:
- weight is
+W - drag is
−F_d - Fᵣₑₛᵤₗₜₐₙₜ = W - F_d
B. Why acceleration decreases
From Newton’s second law:
Fᵣₑₛᵤₗₜₐₙₜ = ma
At the start, F_d is small, so Fᵣₑₛᵤₗₜₐₙₜ is large and the object accelerates quickly.
As speed increases, F_d increases, so W-F_d becomes smaller. This makes the acceleration smaller.
C. Terminal velocity (balanced forces)
Eventually the drag becomes large enough that:
F_d = W
So the resultant force is zero. When Fᵣₑₛᵤₗₜₐₙₜ = 0, acceleration is zero and the object continues at constant velocity (terminal velocity).
See also: Balanced Forces and Newton’s First Law and Unbalanced Force.
D. Velocity–time graph (typical shape)
The graph below is schematic (shape only).
Falling with air resistance (schematic)
Schematic velocity–time graph for a falling object with air resistance, approaching a constant terminal velocity.
Scroll across the graph to read all labels.
View figure data
| Time (arbitrary units) | v–t |
|---|---|
| 0 | 0 |
| 1 | 12 |
| 2 | 22 |
| 3 | 30 |
| 4 | 36 |
| 5 | 41 |
| 6 | 45 |
| 7 | 48 |
| 8 | 50 |
| 9 | 50 |
| 10 | 50 |
Key links to remember:
- gradient of a v–t graph is acceleration
- “flattening” means acceleration is decreasing
- horizontal line means acceleration is zero (terminal velocity)
See: Reading Kinematics Graphs.
E. Parachutes (two terminal velocities)
When a parachute opens, surface area increases, so drag increases suddenly:
- drag can become greater than weight, so the resultant force becomes upwards
- the skydiver is still moving downwards, so this upward resultant force causes a deceleration
- a new, smaller terminal velocity is reached when drag again equals weight
4. Common Mistakes
- Saying “acceleration is always g” even when air resistance is present.
- Saying “terminal velocity means zero velocity” (it means zero acceleration).
- Forgetting that terminal velocity occurs when forces are balanced, not when there are no forces.
- Mixing up the forces: weight acts downwards; drag acts upwards while falling.
5. Exam Tips
- Always start with a quick force statement: “weight down, drag up”.
- Use: Fᵣₑₛᵤₗₜₐₙₜ = W-F_d (for downward motion, taking downward as positive).
- If the question says “constant velocity” or “terminal velocity”, write:
- resultant force = 0 → F_d = W
- For parachute questions, include the key idea: “drag increases because surface area increases”.
6. Worked Examples
- “ignore air resistance”: F_d ≈ 0 so a ≈ g.
- “terminal velocity” / “constant velocity”: a = 0 so F_d = W.
- Use the value of g given (or 10 N kg⁻¹ if not stated).
Modelled example 1
Forces at terminal velocity
Problem
Study the worked solution
Translate constant velocity into acceleration
Method
Set the acceleration to zero.Reason
Constant velocity means neither speed nor direction is changing.Working
a = 0 ⇒ Fᵣₑₛᵤₗₜₐₙₜ = ma = 0 NCalculate and balance the weight
Method
Find the weight, then match it with the upward drag.Reason
Zero resultant requires the two vertical forces to be equal and opposite.Working
W = mg = (80)(10) = 800 N Therefore, air resistance is 800 N upwards.
Guided practice 2
Acceleration while falling (drag smaller than weight)
Problem
Find the resultant before dividing by mass
Hints
Hint 1: choose a positive direction
Hint 2: form the resultant
View solution step by step
Find the resultant force
Method
Subtract upward drag from downward weight.Reason
Forces in opposite directions combine with opposite signs.Working
Fᵣₑₛᵤₗₜₐₙₜ = 20-6.0 = 14 N downwardApply Newton's second law
Reason
Acceleration is resultant force divided by mass.Working
a = 14/2.0 = 7.0 m s⁻² downward
Common misconception 3
After a parachute opens (drag greater than weight)
Learner response
Separate velocity from acceleration
View solution step by step
Calculate the upward resultant
Method
Subtract weight from drag.Reason
Drag is the larger force at this instant.Working
Fᵣₑₛᵤₗₜₐₙₜ = 1200-800 = 400 N upwardFind and interpret acceleration
Method
Divide by mass, then compare acceleration with the existing velocity.Reason
Acceleration changes velocity; it does not set the direction of motion instantly.Working
a = 400/80 = 5.0 m s⁻² upward The skydiver is still moving downward but slowing.
Examiner practice 4
Reading the graph idea
Examination question
Link graph gradient to force balance
View solution step by step
Identify the graph feature
1 markMethod
State that the graph is horizontal.Reason
A horizontal velocity–time graph has zero gradient and therefore zero acceleration.Working
gradient = 0 ⇒ a = 0State the force condition
1 markMethod
State that drag equals weight.Reason
Zero acceleration requires zero resultant force.Working
F_d = W
Self-mark with the mark scheme
Compare your response with each mark point. Select a point only when your response contains that evidence.
Self-mark the graph feature and force condition separately.
Challenge 5
Just after release (drag is very small)
Boundary-condition transfer
Reason from the zero-speed instant
Hints
Hint 1: use the stated approximation
Hint 2: identify the remaining force
View solution step by step
Identify the initial resultant
Method
Treat weight as the only significant force.Reason
Drag is negligible at the stated zero-speed instant.Working
Fᵣₑₛᵤₗₜₐₙₜ ≈ W downwardInfer initial acceleration
Method
Use W = mg in F = ma.Reason
Mass cancels when weight supplies the whole resultant.Working
a ≈ g downward, about 9.8 m s⁻² or the value supplied.
7. Mind Stretchers
Mind stretcher 1: Paper vs crumpled paperExtension
A flat sheet of paper and the same paper crumpled into a ball are dropped from the same height.
Which reaches the ground first? Explain using drag and terminal velocity.
Show Answer
The crumpled paper reaches first.
The flat sheet has a larger surface area, so it experiences larger drag. This gives it a smaller terminal velocity, so it falls more slowly.
Mind stretcher 2: Two objects, same shape but different massExtension
Two objects have the same shape and surface area but different masses. They are dropped from rest in air.
Which has the larger terminal velocity? Explain.
Show Answer
The heavier object has the larger terminal velocity.
It has a larger weight, so a larger drag force is needed to balance it. That larger drag is reached at a higher speed, so the terminal velocity is higher.
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Course and syllabus information
- Course
- SEC G3 Physics
- Edition
- SEC G3 Physics 2027