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Free fall and gravitational acceleration

Key idea: Understand free fall near Earth, constant gravitational acceleration, sign conventions and vertical-motion graphs through concise O-Level examples.

  • SEC G3 Physics 2027
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Learning objectives

  • State what speed means
  • State what velocity means, including its direction
  • Calculate average speed from total distance and total time
  • Calculate acceleration as change in velocity divided by time taken
  • State what uniform acceleration means
  • Interpret examples of non-uniform acceleration
  • Plot and interpret displacement–time and velocity–time graphs in one dimension
  • Deduce rest and uniform or non-uniform velocity from a displacement–time graph
  • Deduce rest, uniform velocity and uniform or non-uniform acceleration from a velocity–time graph
  • Use signed area under a velocity–time graph to determine displacement
  • Recall constant free-fall acceleration near Earth as approximately 10 m/s²

1. Definitions

An object is in free fall when gravity is the only force acting on it. Near Earth’s surface, its acceleration is approximately constant and directed downwards:

g ≈ 10 m s⁻²

Use the value printed in the question when one is given, such as 9.8 m s⁻².

Scope of this lesson

This page uses the free-fall model, so air resistance is negligible. For falling through air, including the changing force stages and terminal speed, study Terminal Velocity.

2. Key Ideas

  • Gravitational acceleration points downwards whether the object is moving up, moving down or momentarily at rest.
  • In the free-fall model, g is constant near Earth’s surface.
  • From rest, an object’s downward velocity increases by about 10 m s⁻¹ each second when g = 10 m s⁻².
  • Mass does not change the value of g in this model.
  • A sign tells direction; it does not make the magnitude of speed or acceleration negative.
Free fall with constant downward accelerationA schematic sequence shows a released ball at equal time intervals with increasing gaps and longer downward velocity arrows. Beside it, a straight velocity–time graph has constant positive gradient when downward is chosen as positive.Equal time intervalsreleased: speed = 0laterlaterlaterdownward velocity increasesVelocity–time modeltimedownward velocityconstant gradient = gdownward chosen as positiveair resistance ignored
Scroll diagram horizontally to read all labels.
With air resistance ignored, a falling object has constant downward acceleration. Its spacing increases in equal time intervals, and its velocity–time graph has constant gradient.

3. Detailed Explanations

Choosing signs

State a positive direction and keep it throughout the calculation.

Positive directionAcceleration in free fall
downwardsa = +g
upwardsa = -g

If upwards is positive, a negative final velocity means the object is moving downwards. It does not mean the calculation has failed.

Graphs for constant g

For an object released from rest with downwards positive:

  • the velocity–time graph is a straight line through the origin with gradient g;
  • the displacement–time graph curves and becomes steeper because velocity is increasing;
  • the area under the velocity–time graph gives downward displacement.

For an upward throw with upwards positive, the velocity–time graph has constant negative gradient -g. It crosses the time-axis at the highest point and continues into negative velocity as the object falls.

Revise reading kinematics graphs and the area under a velocity–time graph if these links between shape and motion are unfamiliar.

4. Common Mistakes

  • Treating gravitational acceleration as negative in every coordinate system. Its sign follows the declared positive direction.
  • Confusing a decreasing upward velocity with a downward velocity before the turning point.
  • Applying constant-g equations after air resistance has made acceleration vary substantially.

5. Exam Tips

  • Draw and label the positive direction before assigning signs to displacement, velocity and acceleration.
  • State the model—uniform gravitational field and negligible air resistance—before using constant-acceleration equations.
  • Use the slope of a velocity–time graph to check whether the sign and magnitude of acceleration agree with the motion description.

6. Worked Examples

Modelled example 1

Speed after a drop

Core

Problem

A ball is released from rest. Take downwards as positive and g = 10 m s⁻². Find its velocity after 2.5 s.
Study the worked solution

  1. Apply the declared sign convention

    Method

    Use u = 0 and a = +10 m s⁻².

    Reason

    Downwards is positive and the free-fall acceleration points downwards.

    Working

    Δ v = aΔ t = (10)(2.5) = 25 m s⁻¹

  2. State velocity with direction

    Reason

    The ball starts from rest, so its final velocity equals the calculated change.

    Working

    v = +25 m s⁻¹ = 25 m s⁻¹ downwards

Guided practice 2

Upward throw with signs

About 5 min

Problem

A ball is thrown upwards at 20 m s⁻¹. Take upwards as positive and g = 10 m s⁻². Find its velocity after 3.0 s.

Keep the upward sign convention throughout

Unit: m s^-2
Unit: m s^-1

Hints

Hint 1: assign acceleration sign
With upwards positive, use a = -g.
Hint 2: interpret the final sign
A negative final velocity means downward motion.
View solution step by step

  1. Assign signed quantities

    Method

    Use u = +20 m s⁻¹ and a = -10 m s⁻².

    Reason

    Gravity remains downward while the chosen positive direction is upward.

    Working

    v = u + at

  2. Calculate and interpret

    Reason

    After 3.0 s, gravity has changed velocity by -30 m s⁻¹.

    Working

    v = 20 + (-10)(3.0) = -10 m s⁻¹, so motion is downward

Common misconception 3

Acceleration at the highest point

Find and correct the mistake

Learner response

At the highest point of an upward throw, a student says velocity and acceleration are both zero. Take upwards as positive. Locate the error and state both quantities at that instant.

Distinguish instantaneous motion from velocity change

Unit: m s^-1
Unit: m s^-2

View solution step by step

  1. Separate velocity and acceleration

    Method

    Set instantaneous velocity to zero but keep gravitational acceleration.

    Reason

    Velocity describes motion at the instant; acceleration describes how that velocity is changing.

    Working

    v = 0, a = -g = -10 m s⁻²

Examiner practice 4

Displacement from the graph area

4 marks

Examination question

A ball falls from rest for 4.0 s. Take downwards as positive and g = 10 m s⁻². Use a velocity–time graph to find its displacement. [4 marks]

Show the final velocity and graph area

View solution step by step

  1. Find the final velocity

    1 mark

    Method

    Use the constant free-fall acceleration.

    Reason

    The ball starts from rest and downwards is positive.

    Working

    v = gt = (10)(4.0) = 40 m s⁻¹

  2. Identify the graph shape

    1 mark

    Method

    Draw a straight line from (0,0) to (4.0,40).

    Reason

    Constant acceleration makes velocity increase linearly.

    Working

    The area under the line is a triangle.

  3. Calculate the area

    1 mark

    Reason

    Velocity–time area gives displacement.

    Working

    Δ s = (1/2)(4.0)(40) = 80 m

  4. State direction

    1 mark

    Method

    Report the signed direction.

    Reason

    The entire area lies in the chosen positive downward direction.

    Working

    Displacement is 80 m downwards.

Challenge 5

Two different masses in free fall

Minimal support

Model comparison

A 0.10 kg ball and a 2.0 kg ball are released together from rest in the same uniform field. Neglect air resistance and take g = 10 m s⁻². Find each velocity after 1.5 s and explain the comparison.

Test whether mass enters the kinematics model

Unit: m s^-1 downward
Unit: m s^-1 downward

Hints

Hint 1: use the stated model
Both objects start from rest and experience the same acceleration g.
Hint 2: calculate one velocity
Use v = (10)(1.5); no mass term is required.
View solution step by step

  1. Apply the same free-fall equation

    Method

    Use v = gt for each ball.

    Reason

    In the negligible-air-resistance model, both masses have the same gravitational acceleration.

    Working

    v_(0.10) = v_(2.0) = (10)(1.5) = 15 m s⁻¹ downwards

Further mistakes to diagnose

  • Saying acceleration is zero at the highest point.
  • Changing the positive direction part-way through a calculation.
  • Using + g after defining upwards as positive.
  • Treating a negative velocity as a negative speed.
  • Calling motion “free fall” when air resistance is significant.
  • Mixing the acceleration unit m s⁻² with the gravitational field-strength unit N kg⁻¹ without explaining the quantity.

7. Mind Stretchers

Mind stretcher 1: Compare two instantsExtension

A ball is thrown vertically upwards at 30 m s⁻¹. Take upwards as positive and g = 10 m s⁻². Find its velocity at t = 2.0 s and t = 4.0 s.

Show answer

Use v = u + at with u = +30 m s⁻¹ and a = -10 m s⁻².

At t = 2.0 s:

v = 30 + (-10)(2.0) = +10 m s⁻¹

At t = 4.0 s:

v = 30 + (-10)(4.0) = -10 m s⁻¹

The velocities have equal magnitudes but opposite directions. The ball passes its highest point at t = 3.0 s.

Mind stretcher 2: Explain the highest pointExtension

At the highest point of a vertical throw, the velocity is momentarily zero. Explain why the acceleration is not zero.

Show answer

Velocity describes the motion at that instant; acceleration describes how velocity is changing. Gravity still acts downwards at the highest point, so the acceleration remains approximately 10 m s⁻² downwards. The downward acceleration changes the velocity from upward to downward.

Continue with the next resource in this course.

Course and syllabus information
Course
SEC G3 Physics
Edition
SEC G3 Physics 2027