IGCSE 0625 / Section 1.2 / Core

Falling at a steady rate.

Drop a heavy hammer and a light feather together, with no air in the way, and they land at the exact same instant. Near the Earth's surface every object gains speed at the same fixed rate regardless of mass. That rate has a name and a strict syllabus value: g, exactly 9.8 metres per second, every second.

The Key Idea

Near the Earth's surface, and strictly ignoring air resistance, all objects fall with the same constant acceleration g = 9.8 m/s². The acceleration does not change as the object falls; the velocity does.

SECTION 01

Equal gains, every second.

Release the objects and watch the strobe marks, one for each second. The gaps grow larger because they are speeding up. Toggle the air resistance on and off to prove the ultimate rule: constant acceleration applies to all masses only when in a vacuum.

SECTION 02

What stays the same, and what grows.

time velocity constant gradient = g
The velocity-time graph for free fall is a straight line through the origin: a constant gradient means a constant acceleration.
Constant acceleration, not increasing acceleration

A very common belief is that a falling object accelerates more and more as it drops. Ignoring air resistance, this is completely wrong. The acceleration stays fixed at 9.8 m/s². What increases is the velocity, and it increases at a steady rate. The growing gaps in a strobe photo show rising speed, not rising acceleration.

Worked Example

A stone is dropped from rest. Taking g = 9.8 m/s² and ignoring air resistance, find its velocity after 3.0 s, and state its acceleration at that moment.

Step 1 : Velocity from v = g t v = g t = 9.8 × 3.0 = 29.4 m/s
Step 2 : Acceleration The acceleration is still 9.8 m/s². It has not changed; only the velocity has grown.
Step 3 : Answer velocity = 29 m/s downward; acceleration = 9.8 m/s², constant throughout.
Practice this topic →
Six original Cambridge-style questions.
The acceleration of free fall, why all masses fall together when air resistance is ignored, calculations using g, and keeping mass and weight distinct. Attempt each, then reveal the worked solution.