The jump, worked
70 kg: W = 70 × 9.8 = 686 N
Air resistance 0 N: a = 686 ÷ 70 = 9.8 m/s²
Air resistance 343 N: a = (686 − 343) ÷ 70 = 4.9 m/s²
Air resistance 686 N: a = 0, her terminal velocity
Terminal velocity Why does a falling skydiver stop speeding up?
Foundation and Higher · AQA · 45 minutes
Students vote on what a skydiver's speed does after she jumps, then follow the forces on her: weight alone at first, air resistance growing with speed until it balances her weight, and a parachute that slows her to a new, lower terminal velocity. A skydiver to try on the slide, her velocity–time graph, a vote that shows who's stuck, practice and homework.
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The idea in one minute
A falling object's weight pulls it down: W = m g, with g = 9.8 N/kg near the Earth. The moment a skydiver jumps she is hardly moving, so there is no air resistance yet: her weight is the only force and she accelerates at g, 9.8 m/s². As she speeds up, air resistance grows. The resultant force, weight − air resistance, gets smaller, and so does her acceleration (a = F ÷ m). Once air resistance equals her weight the resultant force is zero: she stops speeding up and falls at a steady speed, her terminal velocity (Newton's first law). Her parachute has a much bigger area, so at the same speed it gives much more air resistance: the resultant is upwards, she slows down, still falling, until air resistance equals her weight again at a new, lower terminal velocity. A heavier object needs more air resistance to balance its weight, so it reaches a higher terminal velocity.
The lesson follows Newton's laws of motion: the same resultant force and F = m a, now with a force that changes as the speed changes. Students read the story on a velocity–time graph (steep, curving, flat; a sharp drop when the parachute opens, then flat again), its gradient being the acceleration, and try a skydiver on the slide: her mass, when the parachute opens and how big it is. It is for Foundation and Higher; slowing down with the parachute open and the stretch lean towards Higher.
By the end, students can
- work out an object's weight with W = m g
- explain why a falling object's acceleration gets smaller as it speeds up
- explain terminal velocity: air resistance equals weight, so the resultant force is zero
- explain why opening a parachute slows a skydiver to a lower terminal velocity
- sketch and read a skydiver's velocity–time graph, its gradient being the acceleration
Key words
- weight
- mass
- gravitational field strength (g)
- air resistance
- drag
- resultant force
- terminal velocity
- velocity–time graph
- gradient
Before this lesson
Builds on Newton's laws of motion: the resultant force, the first law and F = m a. Students should know a = Δv ÷ t and read a velocity–time graph.
Taught in Newton's laws of motion.
The lesson, step by step
45 minutes in total. The stretch is optional.
5 min 1 · Hook · students think
Faster and faster?
With no air, a skydiver would go faster every second: students vote on what really happens before anyone explains.
Say“A skydiver jumps from a plane. With no air at all, she would gain 9.8 m/s every second: after 20 seconds, nearly 200 m/s. What really happens to her speed?”
Ask- What pushes on her as she falls, apart from gravity?
- Does that push stay the same, or change as she goes faster?
Watch forSaying she falls at a steady speed from the start, or that heavier means faster all the way down: keep the answers, the next step needs them.
For the teacher
Give a minute of silent thinking, then vote. Most choose B; some choose A (gravity keeps pulling) or C (she falls at a steady speed at once).
Don't explain yet. Ask what changes as she speeds up: the air pushing on her. The answer, B, comes from the forces in the next step.
Take a vote with Ask the class: A, B or C. Ask two or three to sketch their guess on the graph. Keep the votes for the explain step.
Shorter: Vote once, no sketches (−2 min).
14 min 2 · Explain · you explain
Weight, air resistance, terminal velocity
Weight W = m g; air resistance grows with speed, so the resultant and the acceleration fall to zero at terminal velocity; the parachute; the velocity–time graph; heavier objects; a falling-ball practical.
Say“Two forces: weight pulls her down and stays the same; air resistance pushes up and grows as she speeds up. The resultant decides everything.”
Ask- Just after she jumps, what is the air resistance on her? So what is her acceleration?
- At terminal velocity, are there no forces, or balanced forces?
- When the parachute opens, which force is bigger? Which way is she still moving?
Watch forSaying there are no forces at terminal velocity, and that she goes up when the parachute opens: ask which way she is moving, and which way the resultant points.
Heavier, a higher terminal velocity
Two skydivers of the same size and shape
60 kg: air resistance must reach 60 × 9.8 = 588 N
90 kg: air resistance must reach 90 × 9.8 = 882 N
More air resistance needs more speed: the 90 kg one falls faster
For the teacher
Weight: W = m g, g = 9.8 N/kg near the Earth. Mass in kg is how much stuff; weight in N is a force. A 70 kg skydiver weighs 686 N, all the way down.
The start: she is hardly moving, so there is no air resistance yet. Her weight is the only force: a = F ÷ m = 686 ÷ 70 = 9.8 m/s², which is g.
Speeding up: air resistance (drag) grows with speed. The resultant force is weight − air resistance, so it shrinks, and so does the acceleration. She is still speeding up, by less each second.
Terminal velocity: air resistance has grown to equal her weight. The resultant is zero, so by Newton's first law she keeps the same velocity. Balanced forces, not no forces.
The parachute: a much bigger area gives much more air resistance at the same speed. Air resistance is now bigger than her weight, so the resultant is upwards: she slows down, still falling. As she slows, air resistance falls, until it equals her weight again at a lower terminal velocity, safe to land at.
The velocity–time graph: steep at first (gradient 9.8 m/s²), curving as the acceleration falls, flat at terminal velocity; a steep drop when the parachute opens, then flat again, lower. The gradient is the acceleration; where it slopes down, she is slowing.
The skydiver on the slide: at 75 kg she falls at most 41.8 m/s, then 5.7 m/s with a 25 m² parachute. Heavier: higher terminal velocities; a bigger parachute: a lower one with it open. (It is a model; real skydivers falling flat reach about 55 m/s.)
Heavier objects: at terminal velocity air resistance equals weight, so a heavier object of the same shape needs more air resistance and so a higher speed. Without air, every object would speed up at 9.8 m/s².
A practical: drop a ball bearing into a tall tube of oil with rubber bands every 10 cm, and time it between the bands. Times getting shorter: it is still speeding up; times the same: terminal velocity, v = distance ÷ time. Or drop paper cake cases, stacked 1 to 5 (the same shape, more mass), from about 2 m: the heavier stacks reach a higher terminal velocity.
On the Try it slide, ask for a prediction before every move: a heavier skydiver first, then a bigger parachute. Its line under the picture gives both terminal velocities.
Shorter: Skip the practical slide; say it in a line (−4 min).
10 min 3 · Your turn · students write
Worked, then yours
Each worked example beside a twin students solve: a weight, an acceleration while falling, and slowing down with the parachute open.
Say“Weight first, then the resultant, then divide by the mass. Which way does the resultant point?”
Ask- Which is bigger here, the weight or the air resistance?
- So is she speeding up, steady or slowing down?
Watch forDividing the weight, or the air resistance, by the mass instead of the resultant force.
Weight first: W = m g
A 70 kg skydiver
W = 70 × 9.8 = 686 N
Then theirs
An 85 kg skydiver. Her weight?
Falling: the resultant, then a
70 kg, air resistance 280 N
Resultant: 686 − 280 = 406 N down
a = 406 ÷ 70 = 5.8 m/s²
Then theirs
80 kg, air resistance 384 N. Her acceleration?
Parachute open: slowing down
75 kg: W = 735 N; air resistance 1500 N just after it opens
Resultant: 1500 − 735 = 765 N up
a = 765 ÷ 75 = 10.2 m/s² upwards: she slows, still falling
Then theirs
60 kg, air resistance 1188 N. How fast does she slow?
For the teacher
Draw the skydiver as a box with two arrows each time: weight down, air resistance up, the longer arrow the bigger force. The resultant is the difference, pointing the way of the bigger one.
In the third pair the resultant is upwards while she moves down: she slows. Ask: is she going up? No: her velocity is still downwards, only smaller each second.
The twins check they find the weight first, subtract to get the resultant, and divide by the mass, not the weight.
Work each example aloud, then students do its twin on paper or the board. Open the answers slide once most have done the twins.
Shorter: The first two pairs only (−4 min).
Longer: Each student writes a skydiver for a partner: a mass and an air resistance, speeding up or slowing down (+3 min).
12 min 4 · Who's stuck? · students vote
Vote, then practise
One question on the parachute opening shows who thinks she goes up; then practice in levels, sorting moments of the fall, and a stretch with air resistance as 0.4v².
Say“Choose an answer on your own. Nobody talks until everyone has chosen.”
Ask- Which way is she moving? Which way does the resultant force point?
Watch forMixing up which way she moves with which way the resultant points: an upward resultant on something falling slows it down.
Show your working on paper. Use g = 9.8 N/kg. Find the weight, then the resultant force, before a = F ÷ m. Give forces in N and accelerations in m/s². The stretch is optional.
The question that shows who's stuck
- A skydiver is falling at her terminal velocity. She opens her parachute. What happens in the next second?
In your Classroom, as they answer (an example) - A7
- B13
- C6
- D4
Each wrong answer is a different mistake. While they answer, your Classroom shows who chose which, so you know in seconds whom to help with what.
- A skydiver has a mass of 65 kg. Use g = 9.8 N/kg. What is her weight, in newtons?
- The moment she steps out of the plane she is hardly moving down, so there is no air resistance on her yet. What is her acceleration, in m/s²?
- An 80 kg skydiver is falling and the air resistance on him is 300 N. What is the resultant force on him, in newtons?
- What is his acceleration at that moment, in m/s²? Give it to 2 decimal places.
- A skydiver weighs 700 N. Put each moment of her jump in the group that says what her speed is doing.They sort 6 items into Speeding up and Steady velocity and Slowing down.
- On a skydiver's velocity–time graph, what does the gradient show?
- When her parachute opens, a skydiver's velocity falls from 50 m/s to 8 m/s in 3 s. On average, by how much does her velocity fall each second, in m/s²?
- Two skydivers of the same size and shape, one 60 kg and one 90 kg, jump together and fall flat. Which reaches the higher terminal velocity, and why?
- Why does a bigger parachute give a lower terminal velocity?
- An 80 kg skydiver opens his parachute. Just after, the air resistance on him is 2000 N. Use g = 9.8 N/kg. How quickly is he slowing down at that moment, in m/s²? Give it to 1 decimal place.
- Stretch (Higher): for a 50 kg skydiver falling flat, the air resistance in newtons is about 0.4 × v², where v is her speed in m/s. Use g = 9.8 N/kg. What is her terminal velocity, in m/s?
Support · Draw the skydiver with weight and air resistance arrows, the bigger force the longer arrow. Bigger down: speeding up. Equal: steady. Bigger up: slowing down.
For the teacher
Everyone chooses before anyone explains. Then ask in turn: which force is bigger now? Which way is the resultant? Which way is she moving? So is she speeding up or slowing down?
Questions 1–4 build fluency (weight, the start, a resultant, an acceleration); 5 sorts moments of the fall; 6–7 read a velocity–time graph; 8–9 ask for reasons; 10 is exam-style. The stretch finds a terminal velocity from a rule for air resistance.
Students show their working on paper or the board; the quiz keeps their final answers.
If many chose that she moves upwards, act it out: a ball thrown down into your hand keeps moving down while your hand pushes up and slows it. Then they practise.
Shorter: After the vote, questions 1–5 only (−5 min).
Longer: The stretch, then try the skydiver at 50 kg to see the answer's idea (+8 min).
4 min 5 · Exit · students show
A skydiver, twice
An acceleration from the resultant force, and what happens once air resistance equals weight, show who has the idea.
Say“Weight first, then the resultant. Then: what does a resultant of zero mean?”
Watch forDividing 450 N by 60 kg, the air resistance alone.
Use g = 9.8 N/kg. Find her weight first.
- A 60 kg skydiver is falling and the air resistance on her is 450 N. What is her acceleration, in m/s²?
- Later, the air resistance on her has grown to 588 N. What happens to her speed now?
For the teacher
A 60 kg skydiver falls and the air resistance on her is 450 N. What is her acceleration?
Later the air resistance has grown to 588 N. What happens to her speed now?
Those who divided one force alone, or said she stops, need it again; set the homework.
Longer: Why does a feather fall more slowly than a coin in air, but not on the Moon? Answer with the forces. (+4 min).
One to one: at each moment of the jump, ask the student to draw the skydiver as a dot with two arrows, weight and air resistance, longer or shorter, and to say whether she is speeding up, steady or slowing down before any numbers; let them drive the skydiver on the slide while you only ask what they predict before each move.
Homework: Terminal velocity
Use g = 9.8 N/kg. For each skydiver, draw a box with an arrow for her weight and one for air resistance, the bigger force the longer arrow. Give units.
- A skydiver has a mass of 55 kg. Find her weight.
- A 70 kg skydiver is falling and the air resistance on him is 210 N. Find the resultant force on him, then his acceleration.
- Explain why a skydiver's acceleration gets smaller as she falls faster, although her weight stays the same.
- A 90 kg skydiver is falling at his terminal velocity. What is the air resistance on him? Explain how you know.
- Sketch the velocity–time graph of a skydiver from jumping until after her parachute has opened. Mark where (a) her acceleration is biggest, (b) air resistance equals her weight, (c) the parachute opens.
- A ball bearing is dropped into a tall tube of oil with rubber bands 10 cm apart. The times between bands are 0.55 s, 0.42 s, 0.40 s, 0.40 s and 0.40 s. Find its terminal velocity in m/s, and explain why the first time is longer.
Print the worksheet with working space and homework. Teacher answers come with your Library copy.
Teaching terminal velocity: common questions
Is terminal velocity on Foundation and Higher, and in combined science?
In AQA GCSE Physics (8463) falling objects, terminal velocity and their velocity–time graphs are on both tiers, and marked physics only: AQA's combined science does not have them. Weight, W = m g, is in both. The lesson suits both tiers; slowing down with the parachute open and the stretch lean towards Higher.
Is the ball falling through oil a required practical?
Not one of AQA's for GCSE Physics. It is worth doing for the same skills: timing, repeating, and spotting from the readings when something has stopped speeding up. Wallpaper paste or washing-up liquid work instead of oil; paper cake cases dropped from about 2 m, stacked one to five, show heavier objects of the same shape reaching a higher terminal velocity.
Where do the skydiver simulation's numbers come from?
It is a model: weight m g against air resistance that grows with the square of the speed, set by the area of the body or of the parachute. At 75 kg it falls at most 41.8 m/s, then 5.7 m/s with a 25 m² parachute. Real skydivers falling flat reach about 55 m/s; the simulation's round numbers are for seeing the shape of the graph and how each change moves it. It runs twice as fast as real time.
Do students need to know that air resistance depends on speed squared?
No. At GCSE it is enough that air resistance (drag) gets bigger as the speed gets bigger, and with a bigger area. The stretch gives a rule with v² in the question, for those who want to find a terminal velocity by calculation.
Should students use g = 9.8 or 10 N/kg?
Use what the question gives. AQA gives the value of g whenever it is needed, usually 9.8 N/kg near the Earth's surface; this lesson and its skydiver use 9.8 throughout.
Can I use the lesson without creating an account?
Yes. Read the lesson, print the worksheet and open its slides, with the skydiver to try, without signing in; nothing you draw there is kept. A free teacher account keeps the whole lesson in your Library to change and teach, with the quizzes, the answers and the answer slides.
Curriculum reference: AQA GCSE Physics 8463: 4.5.1.3 (gravity: weight, W = m g), 4.5.6.1.5 (acceleration: objects falling through a fluid, terminal velocity and their velocity–time graphs; physics only), with 4.5.6.2.1 and 4.5.6.2.2 (Newton's first and second laws).
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