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Drag Force (Air Resistance)

Key idea: Explain drag force (air resistance) qualitatively and use forces + energy to describe terminal velocity (A Level Physics).

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Describe normal, frictional, buoyant and viscous forces qualitatively.

1. Definitions (Must Know)

A. Drag force, F_d

Drag force (air resistance) is a resistive force that acts opposite to an object’s motion through a fluid (air or water).

B. Terminal velocity

Terminal velocity is the constant speed reached when drag equals weight, so the resultant force is zero and acceleration is zero.

O Level foundation: Free Fall (g) & Terminal Velocity.

2. Key Ideas (What Earns Marks)

  • Drag acts opposite the direction of motion.
  • Drag increases with speed; it also depends on shape, area, and the fluid.
  • With drag, the acceleration of a falling object is less than g and decreases as speed increases.
  • At terminal velocity:
    • F_d = W = mg
    • resultant force = 0
    • a = 0 and velocity is constant
  • Energy view at terminal velocity: the rate of loss of gravitational potential energy equals the rate of energy transfer to the surroundings (work done against drag).

3. Detailed Explanations

A. Forces on a falling object (with air resistance)

For an object falling through air, the main forces are:

  • weight W = mg (downwards, approximately constant near Earth)
  • drag F_d (upwards, increases with speed)

At the start, speed is low so drag is small. The resultant force is downwards, so the object accelerates downwards.

As speed increases, drag increases, so the resultant force decreases, so the acceleration decreases.

B. Terminal velocity (force explanation)

Terminal velocity is reached when:

F_d = mg ⇒ Fᵣₑₛᵤₗₜₐₙₜ = 0 ⇒ a = 0

The object continues falling at a constant speed.

Velocity–time with terminal velocity (example)

As speed increases, drag increases and acceleration decreases, so the velocity–time graph flattens towards a constant terminal speed.

Scroll across the graph to read all labels.

As speed increases, drag increases and acceleration decreases, so the velocity–time graph flattens towards a constant terminal speed.As speed increases, drag increases and acceleration decreases, so the velocity–time graph flattens towards a constant terminal speed.
Terminal velocity is the flat (constant) part of the v–t graph where drag equals weight, so resultant force and acceleration are zero.
Open full-size graph
View figure data
Values for Velocity–time with terminal velocity (example)
Time, t (s)Speed (with drag)
00
18.5
214.5
318.5
421
522.7
623.8
724.6
825.1
925.5
1025.7
1125.9
1226

C. Terminal velocity (energy explanation)

As the object falls:

  • gravitational potential energy decreases,
  • energy is transferred to the surroundings by work done against drag (mostly heating the air/object).

At terminal velocity, kinetic energy is no longer increasing. The rate of loss of gravitational potential energy equals the power dissipated by drag.

At terminal velocity, F_d = mg, so: P = F_d v = mgv

D. Factors that affect drag (qualitative)

Drag usually increases when:

  • speed increases,
  • cross-sectional area increases,
  • the object is less streamlined,
  • the fluid is “thicker” (higher viscosity/density, depending on model).

4. Common Mistakes

  • Saying “drag disappears at terminal velocity” (it equals weight; it does not vanish).
  • Saying “terminal velocity means no forces act” (forces act, but they balance).
  • Assuming acceleration is always g in air (only true when drag is negligible).
  • Forgetting drag reverses direction when the motion reverses.

5. Exam Tips

  • Draw a quick free-body diagram and label directions.
  • If the object is speeding up while moving down: resultant force is downwards.
  • If the object is slowing down while moving down: resultant force is upwards (drag > weight).
  • At terminal velocity: write the chain explicitly: F_d = mg ⇒ Fᵣₑₛᵤₗₜₐₙₜ = 0 ⇒ a = 0.

6. Worked Examples

Modelled example 1

Falling but slowing down

Core

Problem

A skydiver is moving downward but slowing down. Compare the magnitudes of drag and weight.
Study the worked solution

  1. Separate velocity from acceleration

    Method

    Velocity is downward but acceleration is upward.

    Reason

    An object slowing while moving downward must accelerate opposite to its velocity.

    Working

    vector v: down; vector a: up

  2. Use the resultant-force direction

    Method

    The resultant force is upward.

    Reason

    Newton’s second law makes resultant force parallel to acceleration.

    Working

    ∑ vector F = m vector a

  3. Compare the two forces

    Method

    Drag is greater than weight.

    Reason

    Upward drag must exceed downward mg to produce an upward resultant.

    Working

    F_d-mg > 0 ⇒ F_d > mg

Guided practice 2

Terminal velocity conditions

About 3 min

Problem

An object is falling at terminal velocity. State its resultant force and acceleration.

Try this before viewing the solution

Unit: N
Unit: m s^-2

Hints

Hint 1: start from constant velocity
Terminal velocity means velocity no longer changes; apply ∑ F = ma.
View solution step by step

  1. Balance drag and weight

    Method

    F_d = mg.

    Reason

    Terminal speed occurs when the upward and downward forces balance.

    Working

    ∑ F = mg-F_d = 0

  2. State the acceleration

    Method

    a = 0.

    Reason

    Zero resultant force gives zero acceleration.

    Working

    0 = ma ⇒ a = 0

  3. Distinguish velocity

    Method

    The object still moves downward at constant non-zero speed.

    Reason

    Zero acceleration means unchanged velocity, not necessarily rest.

    Working

    v = vₜₑᵣₘᵢₙₐₗ = constant

Common misconception 3

Vehicle reaching a steady speed (analogy)

Find and correct the mistake

Learner claim

A vehicle has a constant forward driving force while drag increases with speed. A learner says its eventual steady speed occurs because the driving force falls to zero. Diagnose the explanation.

Try this before viewing the solution

Condition at steady speed

View solution step by step

  1. Track the changing resultant

    Method

    The forward resultant decreases as the vehicle speeds up.

    Reason

    Drag increases while the stated driving force remains constant.

    Working

    Fᵣₑₛᵤₗₜₐₙₜ = F_drive-F_d

  2. Identify the steady-state condition

    Method

    Eventually F_d = F_drive.

    Reason

    The forces balance at the limiting speed.

    Working

    Fᵣₑₛᵤₗₜₐₙₜ = 0

  3. Correct the claim

    Method

    The vehicle continues at constant speed with both forces acting.

    Reason

    Zero resultant gives zero acceleration; the driving force has not vanished.

    Working

    a = 0, v = constant

Examiner practice 4

Finding the drag at terminal velocity

2 marks

Examination question

A skydiver of mass 80 kg falls at terminal velocity. Take g = 9.81 m s⁻². Find the drag-force magnitude. [2 marks]

Try this before viewing the solution

Unit: N

View solution step by step

  1. Apply terminal force balance

    1 mark

    Method

    F_d = mg.

    Reason

    Terminal velocity means zero acceleration and zero resultant force.

    Working

    F_d-mg = 0

  2. Calculate

    1 mark

    Method

    F_d = 7.85 × 10² N.

    Reason

    Drag has the same magnitude as the skydiver’s weight.

    Working

    F_d = 80(9.81) = 784.8 N ≈ 785 N

Challenge 5

Acceleration when drag is known

Minimal support

Independent transfer

A 70 kg skydiver is moving downward while drag is 400 N upward. Take g = 9.81 m s⁻². Find the acceleration magnitude and direction.

Try this before viewing the solution

Unit: m s^-2
Acceleration direction

Hints

Hint 1: form the signed resultant
With downward positive, ∑ F = mg-F_d.
View solution step by step

  1. Choose downward positive

    Method

    Weight is positive and drag is negative.

    Reason

    The forces act in opposite directions.

    Working

    ∑ F = mg-F_d

  2. Calculate resultant force

    Method

    The resultant is 286.7 N downward.

    Reason

    Weight 686.7 N exceeds the 400 N drag.

    Working

    70(9.81)-400 = 286.7 N

  3. Calculate and interpret acceleration

    Method

    a = 4.10 m s⁻² downward.

    Reason

    The positive signed result follows the chosen downward axis.

    Working

    a = 286.7/70 = 4.10 m s⁻²

7. Mind Stretchers

Mind stretcher 1: Two terminal velocitiesExtension

A skydiver reaches a terminal velocity before opening a parachute.

After opening the parachute, they reach a new terminal velocity that is much smaller.

Explain why, using forces.

Click here to show/hide answer

Opening the parachute increases drag for any given speed (larger area/less streamlined).

Immediately after opening, drag becomes much larger than weight, giving an upward resultant force, so the skydiver slows down.

They slow until drag decreases to equal weight again, which happens at a lower speed, giving a lower terminal velocity.

Mind stretcher 2: Optional (Enrichment)Extension

A. Drag models (F_d ∝ v and F_d ∝ v²)

Some problems specify a drag model:

  • low-speed / laminar flow: F_d ∝ v
  • higher-speed / turbulent flow: F_d ∝ v²

If the model is not stated, treat drag qualitatively (increases with speed).

Mind stretcher 3: Why is terminal velocity lower in water than in air?Extension

An object falls in air and reaches a terminal velocity. It falls in water and reaches a much smaller terminal velocity.

Explain why using the idea “drag increases with speed”.

Click here to show/hide answer

For an object falling at terminal velocity, drag must balance weight (F_d = mg).

In water, for the same speed, drag is typically much larger than in air (different fluid properties and flow). So the speed required for drag to build up to mg is smaller.

Therefore the terminal velocity in water is lower.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027