Practise explaining and calculating what happens to a falling object, from the moment it’s dropped to the moment it lands. Answer on paper first, then tap to check. Use g = 9.8 N/kg. Questions marked (H) are Higher tier. For a guided lesson first, try the Skydive Challenge. ← Forces topic
📝 Copy into your book
The terminal velocity chain:
1. Weight > air resistance → resultant force downwards → accelerates.
2. Speed increases → air resistance increases → resultant force and acceleration decrease.
3. Air resistance = weight → resultant force = 0 → constant speed = TERMINAL VELOCITY.
4. Parachute opens → bigger surface area → air resistance > weight → resultant upwards → decelerates.
5. Speed falls → air resistance falls until it equals weight → new, LOWER terminal velocity.
Explaining the motion
Q1 (2 marks) Name the two forces acting on a falling skydiver and give the direction of each.
Show solution
Weight – downwards (1). Air resistance (drag) – upwards, opposite to the motion (1).
Q2 (2 marks) Explain why a skydiver accelerates just after jumping from a plane.
Show solution
Her weight is much bigger than the air resistance (1), so there is a resultant force downwards, which makes her accelerate downwards (1).
Q3 (1 mark) Why does the air resistance on the skydiver increase as she falls?
Show solution
Because her speed increases – the faster an object moves through air, the greater the air resistance (1).
Q4 (2 marks) What is meant by terminal velocity?
Show solution
The constant (maximum) velocity of a falling object (1), reached when the air resistance equals its weight, so the resultant force is zero (1).
Q5 (2 marks) A student says, “At terminal velocity there’s no force on the skydiver.” Explain what is wrong with this statement.
Show solution
There are still two forces acting – weight and air resistance (1). They are equal and opposite, so it is the resultant force that is zero, not the forces themselves (1).
Q6 (3 marks) Explain why opening a parachute makes the skydiver slow down.
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The parachute has a large surface area, so the air resistance increases (1). Air resistance becomes bigger than her weight, so the resultant force is upwards (1). The resultant force is opposite to her motion, so she decelerates (1).
Q7 (2 marks) Explain why the skydiver’s terminal velocity with the parachute open is much lower than without it.
Show solution
With the larger surface area, the air resistance is bigger at any given speed (1). So the air resistance becomes equal to her weight at a much lower speed (1).
Q8 (3 marks – H) Two skydivers of the same size and shape jump together. One has a mass of 60 kg and the other 90 kg. Which reaches the higher terminal velocity? Explain.
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The 90 kg skydiver. She has a greater weight (1), so a greater air resistance is needed to balance it (1). Air resistance increases with speed, so she has to fall faster before the forces balance (1).
Q9 (3 marks – H) A skydiver falling at terminal velocity changes from a spread-out position to a head-down dive. Describe and explain what happens to her speed.
Show solution
Her surface area decreases, so the air resistance decreases (1). Her weight is now bigger than the air resistance, so there is a downward resultant force and she accelerates (1). As she speeds up, air resistance rises until it equals her weight again, at a new, higher terminal velocity (1).
Calculations
Q10–Q14: a skydiver has a mass of 70 kg (including equipment).
Q10 (2 marks) Calculate her weight.
Show solution
W = mg = 70 × 9.8 (1) = 686 N (1)
Q11 (2 marks) At the instant she jumps, the air resistance is almost zero. What is her acceleration?
Show solution
Resultant force = weight = 686 N (1), so a = 686 ÷ 70 = 9.8 m/s² (1) – the same as g.
Q12 (3 marks) A few seconds later, the air resistance is 400 N. Calculate (a) the resultant force and (b) her acceleration.
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(a) 686 − 400 = 286 N downwards (1)
(b) a = 286 ÷ 70 (1) = 4.1 m/s² (1) – still accelerating, but less than before.
Q13 (1 mark) What is the air resistance when she is falling at terminal velocity?
Show solution
686 N (1) – at terminal velocity the air resistance equals her weight.
Q14 (3 marks – H) Just after her parachute opens, the air resistance is 1500 N. Calculate the resultant force and her acceleration. What does the answer tell you?
Show solution
Resultant = 1500 − 686 = 814 N upwards (1)
a = 814 ÷ 70 = 11.6 m/s² upwards (1)
She is still moving down but the resultant is upwards, so she is decelerating at 11.6 m/s² – more than g, which is why the parachute opening feels like a jolt (1).
Q15 (3 marks) A student drops a ball bearing into a tall tube of oil. Marks on the tube are 10 cm apart. She times how long the ball takes to fall between each pair of marks.
| Section | Time (s) |
|---|---|
| 1st | 0.40 |
| 2nd | 0.25 |
| 3rd | 0.20 |
| 4th | 0.20 |
| 5th | 0.20 |
(a) How can you tell the ball has reached terminal velocity? (b) Calculate its terminal velocity.
Show solution
(a) From the 3rd section onwards the time for each 10 cm is the same, so the speed is constant (1).
(b) v = s ÷ t = 0.10 ÷ 0.20 (1) = 0.50 m/s (1) – remember to convert 10 cm to 0.10 m.
Using data
Q16 (6 marks) This data shows a skydiver’s velocity during the first 14 s of her jump.
| Time (s) | Velocity (m/s) |
|---|---|
| 0 | 0 |
| 2 | 18 |
| 4 | 32 |
| 6 | 42 |
| 8 | 49 |
| 10 | 53 |
| 12 | 55 |
| 14 | 55 |
(a) Plot a velocity–time graph and calculate her average acceleration in the first 2 s. (b) What is her terminal velocity? (c) Suggest why her acceleration in the first 2 s is less than 9.8 m/s². (d) She stays at terminal velocity for 40 s. How far does she fall in that time?
Show solution
(a) a = Δv ÷ t = 18 ÷ 2 (1) = 9.0 m/s² (1)
(b) 55 m/s (1) – the velocity stops changing
(c) Air resistance is already acting as she speeds up, so the resultant force is less than her weight (1)
(d) s = v × t = 55 × 40 (1) = 2200 m (1)
Extended answer
Q17 (6 marks) A skydiver jumps from a plane, falls for a while, opens her parachute and lands safely. Sketch a velocity–time graph for her jump, then describe and explain her motion from the moment she jumps until just before she lands.
Hint: use the five-step chain in your 📝 box. For each stage, say what happens to the forces AND to her speed.
Show model answer
Graph: starts at zero and rises steeply, then curves and levels off (terminal velocity); drops steeply when the parachute opens, then curves and levels off at a much lower velocity.
- At first her weight is greater than the air resistance, so the resultant force is downwards and she accelerates.
- As her speed increases, the air resistance increases, so the resultant force and her acceleration decrease.
- Eventually air resistance equals her weight, the resultant force is zero, and she falls at terminal velocity.
- When the parachute opens, the larger surface area makes air resistance greater than her weight.
- The resultant force is upwards, so she decelerates; as her speed falls, the air resistance decreases.
- Air resistance becomes equal to her weight again, so she falls at a new, lower terminal velocity until she lands.
Top-band answers cover all six stages in order and explain each one using forces – not just describe the speed.
✅ Add up your marks out of 46 (or out of 37 if you skipped the H questions Q8, Q9 and Q14).
⚠️ Where students lose marks
- Saying “gravity increases” as she falls. Her weight stays the same – it’s the air resistance that changes.
- Saying the parachute makes her go UP. The resultant force is upwards, but she is still moving down – she just slows down.
- “No forces at terminal velocity.” The forces are balanced; the resultant is zero.
- Describing without explaining. For each stage, say what the forces are doing AND what that does to her speed.
- Using weight in F = ma instead of the resultant force (weight − air resistance).
Related: Resultant forces · F = ma · Motion graph activities
