Free fall and solving motion problems
Key idea: Without air resistance, objects near Earth fall with the same constant downward acceleration regardless of mass. A speed–time graph gives acceleration from its gradient and distance from its area.
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Guided lesson
Guided steps
- Read the free-fall acceleration and when air resistance can be ignored.
- Compare the gradient and distances fallen in successive seconds.
- Explain why mass does not change free-fall acceleration.
- Compare drag with weight, then predict the vacuum drop.
- Find final speed, then use graph area for distance.
- Find the fall time and distance using g = 10 m/s².
- Choose a gradient, area or journey-total method.
- Find the three speeds from changes in distance.
- Explain which graph section is steepest and why.
- Turn the motion description into graph sections and areas.
- Use uniform acceleration between readings to find each graph area.
- Find both rates of speed change, distance and average speed.
- Find both accelerations and compare them with free fall.
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ContinueFor a stone dropped from rest near Earth, ignoring air resistance predicts the same gain in speed each second until impact. That constant gravitational acceleration is the free-fall model. With it, and with the gradients and areas of motion graphs, you can tackle these one-direction motion problems, as long as you pick the right starting point: a table, a graph or a calculation.
Free fall near Earth
The acceleration of free fall is constant
An object is in free fall when gravity is the only force that matters on it. Near the Earth’s surface, the acceleration of free fall is constant and approximately 10 m/s², directed downwards. (The measured value is about 9.8 m/s²; this course uses 10 m/s².)
For an object released from rest and moving downwards, an acceleration of 10 m/s² means its speed rises by 10 m/s every second.
| time after release / s | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| speed / m/s | 0 | 10 | 20 | 30 | 40 |
The speed–time graph of free fall
Speed–time graph of a ball falling freely
Speed in m/s against time in s. The points lie on a straight line through the origin: 0 m/s at 0 s, 10 m/s at 1 s, 20 m/s at 2 s and 30 m/s at 3 s.
Scroll across the figure to read all labels.
View figure data
| Time (s) | Falling ball |
|---|---|
| 0 | 0 |
| 1 | 10 |
| 2 | 20 |
| 3 | 30 |
The graph is a straight line through the origin, so the acceleration is uniform. Its gradient is 30 ÷ 3 = 10 m/s².
The area under the line gives the distance fallen. After 2 s it is 1/2 × 2 × 20 = 20 m; after 3 s it is 1/2 × 3 × 30 = 45 m. The ball falls 5 m in the first second, 15 m in the next and 25 m in the third: it falls further each second because it is moving faster.
Mass does not change the acceleration
Ignoring air resistance, a 5 kg steel ball and a 0.1 kg marble dropped from rest together from the same height gain speed at the same rate and land together. Near Earth, every object in free fall has the same acceleration, about 10 m/s², whatever its mass.
When the free-fall value applies
The value 10 m/s² applies when air resistance is small enough to ignore. That suits a small, dense object such as a stone, a coin or a steel ball falling a short distance. A feather or a flat sheet of paper is slowed noticeably by the air, so it is not in free fall and its acceleration is less than 10 m/s².
Check your understanding 1
A small steel ball and a table-tennis ball of the same size are dropped together from a height of 2 m. Which is closer to free fall, and which is expected to land first? Explain, then state what would change if the drop took place in a vacuum.
Show answer
The steel ball. For comparable size and shape, air resistance is a smaller fraction of the steel ball’s weight, so its motion is closer to free fall. Air resistance affects the lighter table-tennis ball more relative to its weight; under these conditions the steel ball is expected to land first. In a vacuum both would share the same acceleration and land together. Over this short drop, the difference in landing times may be small.
Worked example 1
A stone dropped from a bridge
Problem
A stone is dropped from rest from a bridge and hits the water 2.5 s later. Air resistance is negligible. Find its speed just before it hits the water and the height it falls.
Show full solution
Identify the acceleration
Method
Use g = 10 m/s².Reason
The stone is in free fall, so its speed rises by 10 m/s every second.Working
Answer
a = 10 m/s²
Find the final speed
Method
Rearrange acceleration = (final speed - initial speed) ÷ time.Reason
The stone starts from rest.Working
Answer
final speed = 0 + 10 × 2.5 = 25 m/s
Find the height
Method
Use the area under the speed–time graph.Reason
The graph is a straight line from 0 to 25 m/s over 2.5 s, so the area is a triangle.Working
Answer
1/2 × 2.5 × 25 = 31.25 m, about 31 m
Check
Working
Answer
After 2 s the stone would have fallen 20 m and after 3 s, 45 m. 31 m lies between.
Check your understanding 2
A stone is dropped from rest and reaches 35 m/s just before it lands. How long did it fall for, and how far did it fall? Air resistance is negligible.
Show the steps Optional working
Show answer
Its speed rises by 10 m/s every second, so it takes 35 ÷ 10 = 3.5 s. The distance is the area under the speed–time graph: 1/2 × 3.5 × 35 = 61.25 m, about 61 m.
Solve a motion problem: table, graph or calculation
A motion problem usually gives you one representation and asks for a quantity. Choose the route that connects them. For a problem with several stages, a quick sketch of the speed–time graph is usually the safest start: it shows every acceleration as a gradient and every distance as an area.
| You are given or asked for | Use |
|---|---|
| total distance and total time | average speed = total distance ÷ total time |
| a starting speed, a final speed and a time | acceleration = (final speed - initial speed) ÷ time |
| a description of several stages, or speeds at many times | draw a speed–time graph: gradient for acceleration, area for distance |
| distances at many times | draw a distance–time graph: gradient for speed |
| an object falling freely | acceleration 10 m/s²: its speed rises by 10 m/s every second |
Check your understanding 3
A runner’s distance from the start line is 0 m, 40 m, 100 m and 130 m at 0 s, 10 s, 20 s and 30 s. Assume steady motion within each interval. Calculate the speed during each ten-second interval.
Show the steps Optional working
Show the steps Optional working
Show answer
The speeds are 40 ÷ 10 = 4.0 m/s, (100 - 40) ÷ 10 = 6.0 m/s and (130 - 100) ÷ 10 = 3.0 m/s. She is fastest from 10 s to 20 s. That section of the distance–time graph has the greatest gradient, 6.0 m/s, so it is the steepest.
Check your understanding 4 Find the fastest interval on the graph
Use the runner’s three interval speeds. During which interval is she fastest? Explain how the distance–time graph shows it and why you must subtract successive distances.
Show answer
The runner is fastest from 10 to 20 s, at 6.0 m/s. That distance–time section has the greatest gradient and is steepest. Use the change from 40 to 100 m, not the total 100 m from the start, because the interval covers only those ten seconds.
From a description to a graph
A cyclist moves at 4.0 m/s for 3.0 s, then accelerates uniformly to 10.0 m/s over the next 2.0 s. Find the acceleration, the distance travelled and the average speed.
- There are two stages, so sketch the speed–time graph: horizontal at 4.0 m/s from 0 to 3.0 s, then a straight line up to 10.0 m/s at 5.0 s.
- Acceleration is the gradient of the sloping section: (10.0 - 4.0) ÷ 2.0 = 3.0 m/s².
- Distance is the area: rectangle 4.0 × 3.0 = 12 m, plus trapezium 1/2 × (4.0 + 10.0) × 2.0 = 14 m, total 26 m.
- Average speed = 26 ÷ 5.0 = 5.2 m/s.
- Check: 5.2 m/s lies between 4.0 m/s and 10.0 m/s, closer to 4.0 m/s because the cyclist spends longer at that speed.
From a table to a graph
The table gives a cyclist’s speed every 2 s. Assume uniform acceleration between readings where the speed changes.
| time / s | 0 | 2 | 4 | 6 | 8 | 10 |
|---|---|---|---|---|---|---|
| speed / m/s | 0 | 4 | 8 | 8 | 8 | 0 |
Speed–time graph of a cyclist
Speed in m/s against time in s. The points are 0 m/s at 0 s, 4 m/s at 2 s, 8 m/s at 4 s, 8 m/s at 6 s, 8 m/s at 8 s and 0 m/s at 10 s, joined by straight lines.
Scroll across the figure to read all labels.
View figure data
| Time (s) | Cyclist |
|---|---|
| 0 | 0 |
| 2 | 4 |
| 4 | 8 |
| 6 | 8 |
| 8 | 8 |
| 10 | 0 |
- From 0 to 4 s the speed rises by 4 m/s every 2 s: 8 ÷ 4 = 2.0 m/s².
- From 4 s to 8 s the speed stays at 8 m/s: uniform speed.
- From 8 s to 10 s the speed falls from 8 m/s to 0: acceleration = (0 - 8) ÷ 2 = -4.0 m/s², a deceleration of 4.0 m/s².
- Distance is the area under the graph: 1/2 × 4 × 8 + 4 × 8 + 1/2 × 2 × 8 = 16 + 32 + 8 = 56 m.
- Average speed = 56 ÷ 10 = 5.6 m/s.
The table gave the accelerations directly; the graph made the area, and so the distance, easy to find.
Check your understanding 5
A train accelerates uniformly from rest to 20 m/s in 40 s, travels at 20 m/s for 60 s, then decelerates uniformly to rest in 20 s. Find (a) its acceleration, (b) its deceleration, (c) the total distance and (d) its average speed.
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Show the steps Optional working
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(a) 20 ÷ 40 = 0.50 m/s². (b) 20 ÷ 20 = 1.0 m/s². (c) Area = 1/2 × 40 × 20 + 20 × 60 + 1/2 × 20 × 20 = 400 + 1200 + 200 = 1800 m. (d) 1800 ÷ 120 = 15 m/s.
Check your understanding 6 Does the measured motion fit free fall?
A falling object is released from rest. Its recorded downward speeds are 0, 3 and 5 m/s at 0, 0.5 and 1.0 s. Find the average acceleration in each half-second interval. Do the values fit the near-Earth free-fall model with negligible air resistance? Choose whether the readings fit the model.
Show the steps Optional working
Show answer
The average accelerations are (3 − 0) ÷ 0.5 = 6 m/s² and (5 − 3) ÷ 0.5 = 4 m/s² downwards. They are unequal and below the expected approximately 10 m/s², so these records do not fit the stated free-fall model. Air resistance could matter, but the data alone do not distinguish it from measurement error or other forces. Check the method and conditions rather than declaring a cause from the values alone.
Common mistakes
- Saying heavier objects fall with a greater acceleration. In free fall every object near Earth accelerates at about 10 m/s², whatever its mass.
- Saying the acceleration grows as the object falls faster. The speed grows; the acceleration stays constant.
- Saying 10 m/s² means the object falls 10 m every second. It is a change in speed each second: from rest the object falls 5 m, then 15 m, then 25 m in successive seconds.
- Using 10 m/s² for a falling sheet of paper. Air resistance on it is not negligible, so it is not in free fall.
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Syllabus and review details
- SEC G2 Science Physics component 2027 · 2027
Physics section, PDF pages 8–23
Last reviewed: