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Kinematics · Subtopic 3 of 5

Vertical Motion under Gravity

Everything falls with the same acceleration, g=10 m s−2g=10\ \text{m s}^{-2} downwards. Choose up as positive, set a=−10a=-10, and suvat does the rest, including the moment the ball hangs still at the top.

Cambridge 9709 Paper 4Edexcel 9MA0 Paper 3AQA 7357 Paper 2≈ 35 minRevision 6 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • Model a falling or thrown object as a particle with acceleration gg downwards.
  • Use v=0v=0 at the greatest height, and s=0s=0 (or a negative ss) when it lands.
  • Find the times when the object is at a given height, and how long it spends above it.
  • Solve problems with two objects moving vertically.
Pause and think

A ball is thrown straight up. At the very top, what is its velocity? What is its acceleration?

Section 2 of 14

Core idea

Start hereUp, pause, down

Throw a ball straight up at 20 m/s. It slows, stops for an instant at the top, then falls back. The whole trip has one constant acceleration, gg downward, so the suvat equations handle all of it in one go.

The big idea

Free vertical motion: take up as positive, so a=−10a=-10 m s−2^{-2}. At the top, v=0v=0. A displacement below the start is negative.

1. Thrown upwards

Height of a ball thrown up at 20 m/s
123451015200t (s)height (m)s = 20t − 5t²vtop: v = 0
The arrow is the velocity: steep upwards at first, flat at the top (t=2t=2), steep downwards at the end. It lands at t=4t=4.
Worked exampleThrown up at 20 m/s

Find the time to the top, max height, time to return, return speed.

Goal
Four values.
Strategy
u=20u=20, a=−10a=-10.
  1. 0=20−10t⇒t=2 s0=20-10t\Rightarrow t=2\text{ s}
    Top: v=0v=0.
  2. 0=400−20s⇒s=20 m0=400-20s\Rightarrow s=20\text{ m}
    v2=u2+2asv^2=u^2+2as.
  3. 0=20t−5t2⇒t=4 s0=20t-5t^2\Rightarrow t=4\text{ s}
    Back to s=0s=0.
  4. v=20−40=−20v=20-40=-20
    20 m/s downward.
Answer22 s, 2020 m, 44 s, 2020 m/s down.
✓ CheckSymmetry: up time = down time ✓.

2. From a height

Worked exampleOff a cliff

Thrown up at 20 m/s from a 25 m cliff. When does it hit the sea, and how fast?

Goal
Time, speed.
Strategy
s=−25s=-25.
  1. −25=20t−5t2⇒t2−4t−5=0-25=20t-5t^2\Rightarrow t^2-4t-5=0
    Rearrange; ÷5\div5.
  2. (t−5)(t+1)=0⇒t=5(t-5)(t+1)=0\Rightarrow t=5
    Reject negative time.
  3. v=20−50=−30v=20-50=-30
    v=u+atv=u+at.
Answer55 s; 3030 m/s.
✓ Checkv2=400+500=900v^2=400+500=900 ✓.
Edge caseTwo times for one height 15=20t−5t215=20t-5t^2 gives t=1t=1 (going up) and t=3t=3 (coming down). Both are real.
Common misconceptionMixed signs If up is positive, gg is −10-10 and depths are negative. Stay consistent.
Pause and think

A stone dropped down a well hits water after 3 s. Depth?

Summary

  • a=−ga=-g (up positive).
  • Top: v=0v=0.
Key words
Free fall
Motion under gravity alone.
A ball thrown upwards at 14.7 m s⁻¹ (upwards positive)
0.511.522.53−15−10−5510150time t (s)velocity v (m s⁻¹)v = 14.7 − 9.8ttop: v = 0 at t = 1.5 sback down at 14.7 m s⁻¹
The gradient is −9.8-9.8 throughout, going up, at the top and coming down. The velocity is zero for an instant at the highest point, but the acceleration is not.
HistoryThe hammer and the feather. Galileo is said to have dropped two balls of different mass from the Leaning Tower of Pisa; historians doubt it happened, but his argument that they land together was right. On 2 August 1971, Apollo 15 astronaut David Scott dropped a hammer and a falcon feather on the Moon, where there is no air. They hit the ground at the same moment, live on television.
Fun factg is not quite constant. gg is about 9.78 m s−29.78\ \text{m s}^{-2} at the equator and 9.839.83 at the poles, because the Earth spins and bulges at the middle. On the Moon it is about 1.61.6, and on Jupiter about 2525. Cambridge rounds to 1010 to keep the arithmetic friendly; Edexcel and AQA use 9.89.8.
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