The big idea: Free fall means gravity is the only force acting — air resistance is ignored.
Every object in free fall has the same acceleration: g = 9.81 m s⁻², always pointing down.
Thrown up, dropped, or falling back — the acceleration is g the whole time.
Spot it: pick a direction first: Choose up = positive. Then g is negative (it points down): a = −9.81 m s⁻².
At the highest point the velocity is zero for an instant — but the acceleration is still 9.81 m s⁻² downward.
Free fall is just constant-acceleration motion with a = g. So you use the same suvat equations — the constant-acceleration equations of motion in s, u, v, a, t — with a set to g.
- final velocity (m s⁻¹)
- initial velocity (m s⁻¹)
- acceleration — here a = −g = −9.81 m s⁻² (m s⁻²)
- time (s)
- displacement — how far it falls (m)
- initial velocity (m s⁻¹) — zero if simply dropped
- acceleration of free fall = 9.81 m s⁻² (m s⁻²)
- time of fall (s)
g is a constant — it is given: g = 9.81 m s⁻² is printed on the data booklet's constants page, so you never have to remember the number.
The mass does not matter — a heavy and a light object fall at the same rate (no air resistance).
Worked example — how fast after a drop
A stone is dropped (released from rest) and falls freely for 2.0 s. How fast is it moving just before it lands? Take g = 9.81 m s⁻².
Solution
- Start with the given formula:
- Dropped from rest so u = 0; it falls, so a = g = 9.81 (taking down as positive here):
- Work it out — keep the unit:
Final answer
v ≈ 20 m s⁻¹ downward.
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How this is tested: Free fall is tested in two forms: a quick Paper 1A multiple-choice question, and a fuller Paper 2 calculation.
- What they ask: an object thrown up and returning to its start, or dropped — find the initial speed, the time, or the maximum height. - The classic trap: forgetting the sign. Up is positive, so a = −9.81 m s⁻², and the velocity is negative on the way down.
Up–down symmetry: the time up to the top equals the time down — so total flight time is twice the time to the top.
Returning to the start: If a ball is thrown up and comes back to the same height, its displacement is zero (s = 0).
So at the moment it lands, v = u + at gives the landing velocity = −u: same speed, opposite direction.
IB-style question — find the initial speed
A ball is thrown straight up from someone's hand. It returns to the same height 3.2 s later. Find the speed it was thrown with. Take g = 9.81 m s⁻² and up as positive.
Solution
- By up–down symmetry, the time to the top is half the flight: ttop = 3.2 ÷ 2 = 1.6 s.
- Start with the given formula:
- At the top the velocity is zero (v = 0), and a = −g:
- Rearrange for u — keep the unit:
Final answer
u ≈ 16 m s⁻¹ — the speed it was thrown with (and the speed it lands at).