What this covers
Within Force and Motion I, this standard is one students often meet in structured questions. A focused lesson turns "I understand it" into "I can score it".
How it is examined
It can appear in Paper 1 (objective) and Paper 2 (structured), and where an experiment applies, in Paper 3. We do not predict which questions appear; we prepare the technique for all of them.
A common mistake
How to study it
Learn the definition precisely, practise one or two SPM-style questions with full working, and link it to the rest of Force and Motion I. If it keeps costing marks, a one-to-one lesson fixes exactly that.
Force and Motion I · Formulas · Exam Papers
What you need to know
Free fall is the motion of an object falling under gravity alone, with air resistance assumed negligible. Near the Earth's surface, every freely falling object experiences the same constant acceleration due to gravity, g = 9.81 m s⁻², directed vertically downward, regardless of its mass.
This means a coin and a sheet of paper fall at the same rate in a vacuum, even though air resistance makes the paper fall more slowly in ordinary air. Because g is constant, the four equations of motion apply directly to free fall by substituting a = g, taking downward as positive, or a = −g, taking upward as positive.
The value of g can be found experimentally by timing the fall of a small dense object, such as a steel ball, released from rest through a measured height and timed electronically to reduce human reaction error. A ticker-tape timer can also record the motion of a falling tape, and the spacing between dots is used to find velocity at different points, from which g is obtained as the gradient of a velocity-time graph.
Worked example
Question: A steel ball is released from rest and falls freely for 0.5 s. Calculate its velocity just before landing and the height it has fallen, taking g = 9.81 m s⁻².
Given: u = 0 m s⁻¹, a = g = 9.81 m s⁻², t = 0.5 s. Using v = u + at: v = 0 m s⁻¹ + (9.81 m s⁻² × 0.5 s) = 4.9 m s⁻¹ (to 2 significant figures).
Using s = ut + ½at²: s = (0 m s⁻¹ × 0.5 s) + ½ × 9.81 m s⁻² × (0.5 s)² = 0 m + (½ × 9.81 × 0.25) m = 1.2 m (to 2 significant figures). The ball lands with a velocity of about 4.9 m s⁻¹ after falling a height of about 1.2 m.
This example shows that u = 0 for an object released from rest, which is a detail frequently overlooked.
How it is examined
Paper 1 objective items may ask candidates to state the value and direction of g, or to identify which quantity remains constant during free fall. Paper 2 structured questions commonly require candidates to calculate velocity, displacement or time for a freely falling object using the equations of motion with a = g, or to explain why a feather and a coin fall at different rates in air but at the same rate in a vacuum.
Paper 3 practical work often involves determining g experimentally, requiring candidates to describe the method, identify sources of error such as air resistance and reaction time, and calculate g from a gradient.
Common mistakes include forgetting that u = 0 when an object is released from rest, using the wrong sign convention for direction, mixing up g = 9.81 m s⁻² with the newton unit, and omitting units when stating the final answer. Since no formula sheet is provided, candidates must remember that a = g applies only when air resistance is negligible, and must state this assumption when required.
Source: DSKP KSSM Physics Form 4 and 5 (Versi English) (Bahagian Pembangunan Kurikulum (BPK), KPM)