K326 / K327 / 2027
Kinematics overview

Lesson 5 of 6

Free fall

  • I can state that free-fall acceleration near Earth is constant and approximately 10 m/s^2.Syllabus

    Syllabus K326 / K327, 2(i). State that free-fall acceleration near Earth is constant and approximately 10 m/s^2.

Make a guess. You are not marked.

A hammer and a feather are released together from the same height inside a chamber with all the air pumped out. Which reaches the floor first?

  1. They land at the same time.
  2. The hammer, because it is heavier.
  3. The feather, because it is lighter.
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They land at the same time.

With no air, gravity is the only force on each object, and both accelerate at the same g regardless of mass. Same start, same acceleration, same landing time.

An object is in free fall when gravity is the only force acting on it.

Use a = (v - u) / t to connect the initial velocity, final velocity and elapsed time. Choose a positive direction before assigning signs.

Near the Earth's surface, the acceleration of free fall is approximately constant at 10 m/s2 downwards. It is represented by g. Use this model when air resistance is negligible (small enough to ignore), and use the value of g given in the question.

A ball released from rest gains about 10 m/s of downward velocity each second. Its velocity is about 10 m/s downwards after 1 s, and 20 m/s downwards after 2 s, provided it has not yet hit the ground.

Worked example

A ball dropped from rest

A ball falls for 0.60 s near Earth. Ignore air resistance and take downwards as positive, with g = 10 m/s2. Find its velocity.

  1. Initial velocity: u = 0 because the ball is released from rest.
  2. Rearrange a = (v - u) / t: v = u + at.
  3. Substitute: v = 0 + (10 x 0.60) = 6.0 m/s downwards.

The ball gains 6.0 m/s of downward velocity in 0.60 s. Its acceleration remains 10 m/s2; g is an acceleration, not a velocity.

Try it step by step

Work it out in steps: Change in velocity g x t in m/s, then Final velocity v = u + gt in m/s.

  1. Problem 1. A stone is dropped from rest and falls for 1.2 s. Ignore air resistance and take g = 10 m/s2 downwards. Find its velocity after 1.2 s.
  2. Problem 2. A ball is thrown straight down at 3.0 m/s and falls for 2.0 s. Ignore air resistance and take g = 10 m/s2. Find its velocity after 2.0 s.
  3. Problem 3. A coin is dropped from rest and falls for 0.80 s. Ignore air resistance and take g = 10 m/s2. Find its velocity after 0.80 s.
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Problem 1.

Change in velocity g x t in m/s: g x t = 10 x 1.2 = 12 m/s

Final velocity v = u + gt in m/s: v = u + gt = 0 + 12 = 12 m/s downwards

Problem 2.

Change in velocity g x t in m/s: g x t = 10 x 2.0 = 20 m/s

Final velocity v = u + gt in m/s: v = 3.0 + 20 = 23 m/s downwards

Problem 3.

Change in velocity g x t in m/s: g x t = 10 x 0.80 = 8.0 m/s

Final velocity v = u + gt in m/s: v = 0 + 8.0 = 8.0 m/s downwards

Free fall does not mean no force. Gravity still acts. Two objects released together from rest at the same height have the same acceleration in this model, even if their masses differ. Real air resistance can make their motions different.

Check your understanding

Two small balls of different masses are dropped from rest at the same height. If air resistance is negligible, which explanation is correct?

  1. The heavier ball gains speed faster because it is heavier.
  2. Both move at a constant speed because there is no air resistance.
  3. Both gain about 10 m/s of downward velocity each second.
Show the answer

Both gain about 10 m/s of downward velocity each second.

Both have approximately the same constant downward acceleration. Released together from rest, they fall together in this model.

Optional: measuring g with light gates (not needed for the exam)

Measure free-fall acceleration with two light gates

A short opaque flag passes through two stationary light gates as a compact body falls. Each gate times how long the same known flag length blocks its beam. This gives two speeds, which can be compared over their matching time interval.

Only the flag crosses the beams

Two fixed light gates time a short flag on a freely falling bodyAn oblique apparatus view shows two U-shaped light gates mounted on a fixed support, one below the other. Their forks open towards the viewer. The flag's vertical path lies between each emitter and receiver, in the plane of both beams. The opaque flag is 20.0 millimetres long in the downward direction. A short spacer at the middle of the flag extends forwards, through the open mouths of the gates, to a compact body that remains in front of the gate frames and outside their beam plane. Thus the body cannot add another beam interruption, and the spacer meets the beam plane only within the flag's measured vertical extent. The dashed flag path and blue downward arrow indicate vertical travel without contact. Separate fixed cables join both gates to one electronic timer or logger. A padded catch below both measured intervals receives the body. The setup is schematic; no measured gate separation is supplied or needed.Gate 1Gate 2Timer /loggerL = 20.0 mmFlagCompact bodyin frontDownwardtravelLight beamPaddedcatch
The body is forward of the beam plane, with clearance from both gate frames. The spacer attaches within the flag's measured height. Only the flag determines each blocked duration; the catch is below the measured region.
  1. Measure the flag length parallel to its travel. Align the same flag with both beams and secure the gates. The body and attachment must neither add another interruption nor strike a gate.
  2. Connect the gates to a compatible timer or logger. Select a mode that records each blocked duration and the times of those intervals on the same clock.
  3. Release the body above the measured region so that it travels vertically without appreciable rotation. Use the catch below. Record several falls to inspect the spread of the readings.

Use a supplied set of readings

In this example, L = 20.0 mm = 0.0200 m. The length instrument resolves 0.1 mm and the logger resolves 0.0001 s. The clock's zero is the start of the first blockage; the body is already moving then.

Both gates time the same flag in one fall. Downward is positive.
ReadingGate 1Gate 2
Block begins / s0.00000.1070
Block ends / s0.02000.1170
Blocked duration / s0.02000.0100
Midpoint time / s0.01000.1120
Average speed during blockage / m/s0.0200 / 0.0200 = 1.000.0200 / 0.0100 = 2.00

Length divided by blocked duration gives the average speed during that short interval. The body keeps moving down, so these are also positive downward velocities. For uniform acceleration, each interval's average velocity equals the velocity at its time midpoint.

Match the interval to the two speed readings

Both rows share one linear time scale. Shaded windows show the blocked intervals; the dots mark their midpoints.

The two midpoint speeds are separated by 0.1020 secondsOne shared linear time axis runs from zero to 0.1200 seconds. Gate 1 is blocked from 0.0000 to 0.0200 seconds, with midpoint 0.0100 seconds. Gate 2 is blocked from 0.1070 to 0.1170 seconds, with midpoint 0.1120 seconds. Its window is exactly half as wide. Dotted vertical guides join the two midpoint positions to a dimension line showing 0.1020 seconds between them. This is the interval matching the change from 1.00 to 2.00 metres per second. It is neither the 0.1070-second start-to-start time nor the 0.0870-second gap between the blocked windows.Gate 10.0200 s blockedGate 20.0100 s blockedt1 = 0.0100 st2 = 0.1120 s0.1020 s0.000.040.080.12Time / s
The first average velocity belongs at t1 = 0.0100 s and the second at t2 = 0.1120 s under the uniform-acceleration model. Their matching interval is 0.1020 s.

Calculate the acceleration

Matching time interval = 0.1120 - 0.0100 = 0.1020 s.

Acceleration = change in velocity / matching time interval
= (2.00 - 1.00) / 0.1020 = 9.80392... m/s2.

Report about 9.8 m/s2 downward, consistent with using 10 m/s2 as the usual approximate value. The gate separation is not needed for this flag-timing calculation.

The 0.1070 s between the two starts, and the 0.0870 s gap between the blockages, describe different intervals. Neither matches these midpoint velocities.

Choose instruments and check the model

  • Length: choose an instrument whose range covers the whole flag and whose resolution supports a 20.0 mm reading. For these supplied readings it resolves 0.1 mm. Check its zero and measure along the travel direction.
  • Timing: use a compatible electronic mode and range that capture the whole event and resolve 0.0100 s blockages. The example resolution is 0.0001 s. A hand-operated stopwatch cannot usefully time such a short interruption.
  • Alignment and forces: keep the flag aligned, avoid rotation or contact, and use a compact body and short flag to limit air resistance. Check that the effective optical interruption length agrees sufficiently with the measured length.

Call the measured acceleration g only while gravity is the only significant force after release. Air resistance or contact can invalidate the free-fall interpretation. In less uniform motion, a short-interval average is only an approximation to a local velocity; extra timer digits do not make it an exact instantaneous value.

Repeated falls reveal variation, but do not correct a wrong flag length, systematic beam misalignment or a wrongly chosen time interval. Improve the cause of the error; timing resolution alone is not accuracy.

Check your understanding

Two short light-gate intervals give downward speeds of 1.00 and 2.00 m/s. Their midpoint times are 0.0100 and 0.1120 s. For approximately uniform acceleration, which calculation uses the correct time interval?

  1. (2.00 - 1.00) / 0.0100, giving 100 m/s2 downwards.
  2. (2.00 - 1.00) / 0.1020, giving about 9.8 m/s2 downwards.
  3. (2.00 - 1.00) / 0.1070, using the time between the starts of the blockages.
Show the answer

(2.00 - 1.00) / 0.1020, giving about 9.8 m/s2 downwards.

The velocities belong to the interval midpoints, so their matching separation is 0.1120 - 0.0100 = 0.1020 s. Divide the change in velocity by that interval.

Practise an exam question

Write your answer first, then tick the mark points it makes.

Exam-style question

A stone is dropped from rest from a bridge and hits the water 1.5 s later. Air resistance is negligible and the acceleration of free fall is 10 m/s2. Calculate the speed of the stone as it hits the water and the height of the bridge above the water. [3]

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