The big idea: Drop a ball, or throw one up — gravity pulls it down the same way. In free fall (gravity only, no air resistance) every object speeds up at the same rate: g = 9.81 m s⁻² downward.
Thrown up, dropped, or falling back — the acceleration is g the whole time, whatever the mass.
Snapped at equal times: the ball slows going up, stops at the top, then speeds up coming down.
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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.
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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).
A stone is dropped (released from rest) and falls freely for 2.0 s. How fast is it moving just before it lands? (g = 9.81 m s⁻²)
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How this is tested — free fall is just suvat with a = g. It comes up two ways:
Paper 1A
- A quick multiple-choice.
- e.g. 'the acceleration at the top?'
Paper 2
- A fuller calculation.
- thrown up / dropped → find speed, time, or max height.
The two tricks that crack these: Sign: take up = +, so a = −9.81; velocity goes negative coming down.
Up–down symmetry: time up = time down, so total flight = 2 × time to the top — and it lands at the same speed it left. Back to the same height → displacement s = 0.
A ball is thrown straight up and returns to the same height 3.2 s later. (Take g = 9.81 m s⁻², up positive.)
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