Air resistance, drag and terminal velocity

What you need to be able to do

Resistive forces: friction and drag

A resistive force always acts against the motion (against the velocity relative to the surface or the fluid). There are two kinds you need:

Drag on an object is larger when:

• its speed is larger (the one that matters most in questions: as speed changes, drag changes);
• its cross-sectional area facing the flow is larger (an open parachute, a snowflake compared with a raindrop of the same mass);
• the fluid is denser or more viscous (oil compared with water, water compared with air).

Drag on a sphere depends on its size and speed, not on its mass. Two balls of the same size moving at the same speed feel the same drag, whatever they are made of. That one fact answers most "compare two objects" questions.

Falling with air resistance: three stages

Drop an object from rest in air. Only two forces act on it: its weight mg (constant, down) and air resistance D (up, growing with speed). Follow the chain at each stage:

1. At release: v = 0, so there is no air resistance. Resultant = mg, so a = g = 9.81 m s⁻².
2. Speeding up: as v increases, air resistance D increases. Resultant = mg − D gets smaller, so the acceleration a = (mg − D)/m gets smaller. It still speeds up, but more and more slowly.
3. Terminal velocity: D has grown until D = mg. Resultant force = 0, so a = 0 and the velocity stays constant (first law). This constant velocity is the terminal velocity.

On graphs, the same story:

In a liquid there is a third force, the upthrust (chapter 4), which is constant. At terminal velocity: weight = drag + upthrust. The story is otherwise identical.

The skydiver and the parachute

Thrown upwards with air resistance

Now the air resistance points down on the way up (against the motion) and up on the way down. The sign of the drag flips at the top.

Comparing two falling objects

For any "which falls faster / reaches the higher terminal velocity" question, use \(a = g - \dfrac{D}{m}\) and the rule that terminal velocity is reached when D = mg.

How to write a "describe and explain" answer

These answers are marked as a chain of cause and effect: one mark for the change in the resistive force with its cause (the speed), one for its effect on the resultant force and the acceleration. Write the chain in this order every time:

1. Speed changes (increases / decreases) →
2. drag / air resistance changes the same way (it depends on speed) →
3. resultant force changes (say how: weight − drag, or weight + drag) →
4. acceleration changes (a = F/m), and say where it ends (to zero, at terminal velocity).

Mark-scheme focus

Worked examples

Example 1 · Skydiver: terminal velocity, tangent and parachute (Paper 2, 10 marks) ★★ Medium

1 Read it with me what is actually happening?

A skydiver falls; her speed rises quickly at first, then more slowly, until it stops changing at 39 m s⁻¹. Then she opens her parachute: the drag suddenly jumps to 1800 N, she slows down, and finally falls steadily at 5.7 m s⁻¹. Each part zooms into one stage of that story.

• terminal velocitythe constant speed where the graph levels off
• By drawing a suitable linea TANGENT at t = 9.0 s; its gradient is the acceleration
• upthrust negligibleonly two forces: weight down, drag up
• magnitude and directiontwo answers: a number and 'up' or 'down'
• Describe and explainsay what happens (acceleration decreases) AND why (drag falls as speed falls)

2 What is it asking for? the target, part by part

• (a)(i)terminal velocitym s⁻¹ · 1 mark
• (a)(ii)acceleration at 9.0 sm s⁻² · 2 marks
• (b)acceleration when the parachute opensm s⁻² + direction · 3 marks
• (c)(i)how the acceleration changes from t₂ to t₃words · 2 marks
• (c)(ii)change in momentum from t₁ to t₃N s · 2 marks

3 What do we know? pull every number out, in SI units

• m = 68 kgskydiver
• drag = 1800 Nat t₁
• v = 39 m s⁻¹at t₁ (terminal)
• v = 5.7 m s⁻¹at t₃

• at t₁ she is still moving at 39 m s⁻¹ DOWN - the parachute has only just opened

4 Plan the route work backwards from what we want

FindPhysics ideaAlready have itWork out firstTrap

• (a)(ii) Finda at 9.0 s
  • Usegradient of the tangent at 9.0 s
    • Trapa chord from the origin gives an average, not the value at 9.0 s
• (b) Finda at t₁
  • Usea = resultant ÷ m
    • Resultantweight (68 × 9.81) − drag (1800)
    • Directiondrag > weight, so the resultant is UP
• (c)(i) Explaina from t₂ to t₃
  • Chainspeed ↓ → drag ↓ → resultant ↓ → a ↓ (to 0)
• (c)(ii) FindΔp
  • Usem(v − u)
    • Haveu = 39, v = 5.7

At the moment the parachute opens, is she moving up or down? Which way is her acceleration?

Still moving DOWN at 39 m s⁻¹, but the acceleration is UP because drag (1800 N) beats weight (667 N). She is slowing down.

Why does her deceleration get smaller as she slows down?

Drag depends on speed. As she slows the drag falls towards the weight, so the resultant force - and the acceleration - shrinks to zero.

5 Now solve it have a go on paper first, then open the worked solution

1. (a)(i) Where the graph levels off: 39 m s⁻¹ A1
2. (a)(ii) Draw the tangent at t = 9.0 s, as long as the grid allows C1. Gradient, e.g. (44 − 26)/(18 − 0) = 1.0 m s⁻² (0.9 to 1.1 accepted) A1
3. (b) Forces at t₁: weight 68 × 9.81 = 667 N down, drag 1800 N up. Down = +: resultant = 667 − 1800 = −1133 N C1
4. a = −1133 / 68 = −17 → 17 m s⁻² A1, direction upwards B1 She is still moving down; she is slowing down.
5. (c)(i) As speed decreases, the drag decreases B1, so the resultant force decreases and the magnitude of the acceleration decreases (to zero at t₃) B1
6. (c)(ii) Δp = m(v − u) C1 = 68 × (5.7 − 39) = (−)2300 N s A1 At t₁ she is still at 39 m s⁻¹.

17 m s⁻² is bigger than g: an opening parachute is a hard jolt. Check the drag is bigger than the weight before you say "upwards".

6 Check it does the answer make sense?

17 m s⁻² upwards is bigger than g - an opening parachute is a hard jolt, so that is believable. Δp is negative: she lost momentum.

Example 2 · A rock thrown up on another planet (Paper 2, 9 marks) ★★ Medium

1 Read it with me what is actually happening?

A rock is thrown straight up on another planet. Its velocity falls in a straight line, passes zero at the top (2.5 s), and becomes negative as it falls back. No air, so the only force is the rock's weight - but g is NOT 9.81 here.

• on another planetdon't use 9.81 - get the acceleration from the graph
• height at t = 4.0 sdisplacement = area under the v-t graph (above minus below the axis)
• change in momentumvelocity goes from +4.0 to −2.4: the sizes add
• no atmospherethe weight is the ONLY force, so it equals the resultant
• State and explain the variationwith air: what happens to the drag as the rock slows, and so to the resultant

2 What is it asking for? the target, part by part

• (b)(i)height at 4.0 sm · 3 marks
• (b)(ii)change in momentum, 0 to 4.0 sN s · 2 marks
• (b)(iii)weight on this planetN · 2 marks
• (c)how the resultant changes going up (with air)words · 2 marks

3 What do we know? pull every number out, in SI units

• m = 7.5 kgrock
• u = +4.0 m s⁻¹t = 0
• v = 0t = 2.5 s (top)
• v = −2.4 m s⁻¹t = 4.0 s, read off the graph

• the line is straight, so the acceleration (and the force) is constant

4 Plan the route work backwards from what we want

FindPhysics ideaAlready have itWork out firstTrap

• (b)(i) Findheight at 4.0 s
  • Usearea under v-t
    • Uptriangle ½ × 2.5 × 4.0
    • Downtriangle ½ × 1.5 × 2.4
    • TrapSUBTRACT the part below the axis
• (b)(ii) FindΔp
  • Usem(v − u) = 7.5 × (−2.4 − 4.0)
• (b)(iii) Findweight
  • Useonly force = weight = Δp ÷ Δt
    • NeedsΔp from (b)(ii)
• (c) Explainresultant with air, going up
  • Chainspeed ↓ → air resistance ↓; weight constant → resultant ↓

Why is the height at 4.0 s less than the maximum height?

After 2.5 s the rock is coming back down. The area below the axis is that downward motion, so it is taken away.

How can Δp give us the weight?

With no air, the weight is the resultant force, and resultant = Δp ÷ Δt. So W = 48 ÷ 4.0.

5 Now solve it have a go on paper first, then open the worked solution

1. (b)(i) Displacement = area under the v–t graph C1. Read v at 4.0 s: −2.4 m s⁻¹ C1
2. Height = area above the axis − area below = ½ × 2.5 × 4.0 − ½ × 1.5 × 2.4 = 5.0 − 1.8 = 3.2 m A1 Subtract the area below the axis: it is motion downwards. The June 2022 report: some added them; some assumed g = 9.81 on this planet.
3. (b)(ii) Δp = 7.5 × (−2.4 − 4.0) C1 = (−)48 N s A1 Velocity reverses, so the speeds add.
4. (b)(iii) The only force is the weight, so W = Δp / Δt C1 = 48 / 4.0 = 12 N A1 (or W = m × gradient = 7.5 × 1.6)
5. (c) Speed decreases so the viscous force decreases B1; weight constant, so the resultant force decreases B1

6 Check it does the answer make sense?

The weight (12 N) gives g = 12 ÷ 7.5 = 1.6 m s⁻², which matches the graph's gradient (4.0 ÷ 2.5) - two routes agree. A small g like that fits a small planet or moon.

Example 3 · "Explain why this force cannot be air resistance" (Paper 2, 10 marks) ★★★ Hard

Why it is hard: part (e) needs you to turn the graph into a statement about the force, and compare it with what air resistance would do. (No examiner report exists for June 2025.)

1 Read it with me what is actually happening?

We get a momentum-time graph that is a straight line falling from +2.8 through zero (at 8 s) to −1.4. So the object slows down steadily, stops, then moves back the other way. We find Δp, the force, describe the speed, argue why the force can't be air resistance, and sketch the displacement.

• Define momentummass × velocity
• constant massthe graph's shape is the velocity's shape
• resultant forcegradient of the p-t graph
• By reference to Fig. 2.1, explainuse what the GRAPH shows (straight line, non-zero at 8 s) in your reasons
• Sketch the displacementrises while p is positive, peaks where p = 0, then falls

2 What is it asking for? the target, part by part

• (b)change in momentum, 0-12 sN s · 1 mark
• (c)size of the resultant forceN · 2 marks
• (d)how the speed changes, 0-8 swords · 1 mark
• (e)why it can't be air resistancewords · 2 marks
• (f)displacement-time sketchsketch · 3 marks

3 What do we know? pull every number out, in SI units

• p = +2.8 N st = 0
• p = 0t = 8.0 s
• p = −1.4 N st = 12 s

4 Plan the route work backwards from what we want

FindPhysics ideaAlready have itWork out firstTrap

• (c) Findforce
  • Usegradient = Δp ÷ Δt
    • NeedsΔp from (b): −1.4 − 2.8
• (e) Explainnot air resistance
  • Comparewhat the graph shows vs what drag would do
    • Graphstraight line → force constant; force still there at v = 0
    • Dragwould fall as speed falls, and be zero at zero speed
• (f) Sketchd against t
  • Usegradient of d-t = velocity ∝ p
    • Somax d at 8.0 s, then back down

What does a straight line on a p-t graph say about the force?

The gradient is the resultant force, so a straight line means the force is constant.

What would air resistance be when the object is momentarily at rest?

Zero - drag only exists while moving. But the graph keeps changing at 8.0 s, so a force is still acting. It can't be drag.

5 Now solve it have a go on paper first, then open the worked solution

1. (a) "product of mass and velocity" B1
2. (b) Δp = (−1.4) − (+2.8) = (−)4.2 kg m s⁻¹ A1
3. (c) F = Δp/Δt = gradient = 4.2 / 12 C1 = 0.35 N A1
4. (d) The speed decreases at a constant rate to zero B1 (straight line → constant force → constant deceleration).
5. (e) Any two B2: the straight line shows the resultant force is constant; air resistance would decrease as the speed decreases; and at 8.0 s the speed is zero but the force is not zero (air resistance would be zero at zero speed, and the object would not then move back).
6. (f) From the origin with decreasing positive gradient B1; maximum at 8.0 s where the gradient becomes negative B1; then a negative gradient of increasing size, with d still positive at 12 s B1

The test for "is this force air resistance?": air resistance always opposes the motion, changes with speed and is zero at zero speed. A constant force that keeps acting through v = 0 is something else.

6 Check it does the answer make sense?

The force (0.35 N) is constant all the way, which is exactly why the momentum carries on through zero and goes negative.

Example 4 · Distance from a curved v–t graph (Paper 1, 1 mark) ★★ Medium

1 Read it with me what is actually happening?

A parachutist's speed rises along a curve to 60 m s⁻¹, stays there for a while, then drops sharply when the parachute opens. We want roughly how far she fell BEFORE it opened.

• Approximately how fardistance = area under the v-t graph - count squares, estimates are expected
• before she opens the parachutestop at the moment v starts to fall (about 13.5 s), not at the end of the graph

2 What is it asking for? the target, part by part

• distance fallen before the parachute opensm · MCQ

3 What do we know? pull every number out, in SI units

• 0 → 60 m s⁻¹curve, 0 to about 10 s
• 60 m s⁻¹constant, 10 to about 13.5 s

• one large square is 5 s × 20 m s⁻¹ = 100 m of distance

4 Plan the route work backwards from what we want

FindPhysics ideaAlready have itWork out firstTrap

• Finddistance
  • Usearea under v-t up to 13.5 s
    • Curved partcount squares 0-10 s (~3.6 squares)
    • Flat partrectangle 3.5 s × 60 m s⁻¹
    • Trapa triangle under the curve under-counts - it bulges above the straight line

What is one large grid square worth in metres?

Width × height in the graph's units: 5 s × 20 m s⁻¹ = 100 m.

Where do we stop counting?

Where the parachute opens - the moment the velocity starts to fall - about 13.5 s.

5 Now solve it have a go on paper first, then open the worked solution

1. Distance = area under the v–t graph up to the moment the parachute opens (when v starts to fall, about 13.5 s).
2. Value of one large square: 5 s × 20 m s⁻¹ = 100 m.
3. Count the squares under the curve: about 3.6 from 0 to 10 s, plus 3.5 s × 60 m s⁻¹ = 210 m (2.1 squares) from 10 s to 13.5 s.
4. Total ≈ 5.7 × 100 = 570 m → B

Stop counting where the parachute opens (where v starts to fall), not at 20 s. And don't treat 0–10 s as a triangle (300 m): the curve bulges above a straight line, so a triangle under-counts.

6 Check it does the answer make sense?

A triangle 0-10 s would give 300 m, and the curve must be MORE than that; 570 m is the only option that fits a careful count.

Summary

• Friction (solid surfaces) and drag (fluids) oppose motion. Drag is zero at rest and increases with speed (and with area and fluid density).
• Falling from rest: a = g at first; drag grows, resultant falls, a falls; terminal velocity when drag = weight (+ upthrust in a liquid), resultant zero.
• Parachute opens: drag ≫ weight, resultant upwards, slows down; drag falls with speed until drag = weight at a lower terminal velocity.
• Thrown up: drag and weight both down, deceleration > g; coming down, a < g; lands slower than thrown.
• Explain answers: speed → drag → resultant → acceleration, each change with its reason.

Can I do this?

Quiz