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).

  • SEC G3 Physics 2027
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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

What you need for this course

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)
Force changes as a falling object approaches terminal velocityThree stages show the same downward weight arrow. The upward drag arrow is small just after release, larger as speed rises, and equal to weight at terminal velocity. The resultant and acceleration decrease to zero.Falling object: three stagesJust after releaseSpeed increasingTerminal velocitysmall dragweightlarge downward resultantlarger dragweightsmaller resultantdragweightresultant = 0constant velocity
Scroll diagram horizontally to read all labels.
Weight stays approximately constant. Increasing drag reduces the downward resultant until drag equals weight and acceleration becomes zero.

The resultant force while falling is:

Taking downwards as positive:

Fᵣₑₛᵤₗₜₐₙₜ = W - F_d

Sign convention

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.

Schematic velocity–time graph for a falling object with air resistance, approaching a constant terminal velocity.Schematic velocity–time graph for a falling object with air resistance, approaching a constant terminal velocity.
Schematic velocity–time graph for a falling object with air resistance, approaching a constant terminal velocity.
Open full-size graph
View figure data
Values for Falling with air resistance (schematic)
Time (arbitrary units)v–t
00
112
222
330
436
541
645
748
850
950
1050

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

Hidden assumptions to watch for
  • “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

Core

Problem

A parachutist of mass 80 kg descends vertically at a constant velocity of 3.0 m s⁻¹. Take g = 10 N kg⁻¹. Find the resultant force and air resistance.
Study the worked solution
  1. 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 N
  2. Calculate 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)

About 5 min

Problem

A 2.0 kg ball falls through air. Its weight is 20 N and drag is 6.0 N. Find its acceleration, including direction.

Find the resultant before dividing by mass

Unit: m s^-2

Hints

Hint 1: choose a positive direction
Take downward as positive.
Hint 2: form the resultant
Use Fᵣₑₛᵤₗₜₐₙₜ = 20-6.0 before applying F = ma.
View solution step by step
  1. 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 downward
  2. Apply 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)

Find and correct the mistake

Learner response

An 80 kg skydiver has weight 800 N downward and, just after the parachute opens, drag 1200 N upward. A student says: “The skydiver accelerates upward, so they immediately move upward.” Diagnose the error and calculate the acceleration.

Separate velocity from acceleration

Immediate motion

View solution step by step
  1. Calculate the upward resultant

    Method

    Subtract weight from drag.

    Reason

    Drag is the larger force at this instant.

    Working

    Fᵣₑₛᵤₗₜₐₙₜ = 1200-800 = 400 N upward
  2. Find 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

2 marks

Examination question

An object falls through air and eventually reaches terminal velocity. State how the velocity–time graph shows that terminal velocity has been reached, and explain the force condition then. [2 marks]

Link graph gradient to force balance

View solution step by step
  1. Identify the graph feature

    1 mark

    Method

    State that the graph is horizontal.

    Reason

    A horizontal velocity–time graph has zero gradient and therefore zero acceleration.

    Working

    gradient = 0 ⇒ a = 0
  2. State the force condition

    1 mark

    Method

    State that drag equals weight.

    Reason

    Zero acceleration requires zero resultant force.

    Working

    F_d = W

Challenge 5

Just after release (drag is very small)

Minimal support

Boundary-condition transfer

A skydiver steps out of a plane. At the instant their downward speed is zero, air resistance is negligible. State the resultant force and initial acceleration, with directions.

Reason from the zero-speed instant

Hints

Hint 1: use the stated approximation
At zero speed, take drag as approximately zero.
Hint 2: identify the remaining force
Only weight is significant at that instant.
View solution step by step
  1. 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 downward
  2. Infer 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.

Continue with the next resource in this course.

Course and syllabus information
Course
SEC G3 Physics
Edition
SEC G3 Physics 2027