LearnYMA01WME01 · Mechanics 1Kinematics of a particle in a straight lineVertical motion under gravity

Kinematics of a particle in a straight line · Subtopic 2 of 3

Vertical Motion under Gravity

Without air resistance, every object near the Earth accelerates downwards at $g=9.8$ m s⁻². Pick a positive direction (usually up), make $a=-9.8$ , and use suvat. At the top of its flight an object is momentarily still: $v=0$ .

Edexcel YMA01 WME01≈ 35 minRevision 5 · updated 2026-10-27

Section 1 of 14

Learning objectives

Use $a=\pm9.8$ with a clear positive direction.
Find times and speeds for dropped objects.
Find greatest height and time to the top using $v=0$ .
Find when an object is at a given height, including starting above the ground.

Pause and think

A ball is thrown straight up and caught at the same height. Compare the time going up with the time coming down, and the speed thrown with the speed caught.

Section 2 of 14

Core idea

Start hereThrowing a ball straight up

Throw a ball up: it slows, stops for an instant, and falls back faster and faster. The whole flight has the same acceleration, $9.8$ m s $^{-2}$ downwards, so one set of suvat equations describes it from start to finish.

The big idea

Take up as positive: $a=-9.8$ . At the top $v=0$ . Displacement below the start is negative.

1. Up and down

Ball projected up at 14.7 m/s

t (s) = 0.05

Symmetric:

1.5

s up,

1.5

s down. It returns to the start at

14.7

m s

^{-1}

downwards.

Worked exampleProjected up at 14.7 m s⁻¹

Time to the top, greatest height, time to return.

Goal: $t$ , $s$ , total $t$ .
Strategy: $v=0$ at the top.

$0=14.7-9.8t\Rightarrow t=1.5$
$v=u+at$ .
$0=14.7^2-19.6s\Rightarrow s=11.0$
$v^2=u^2+2as$ .
$0=14.7t-4.9t^2\Rightarrow t=3$
Back to $s=0$ .

Answer

1.5

11.0

3

✓ CheckSymmetric: up time = down time ✓.

Worked exampleFrom a 20 m tower

Same projection from the top of a 20 m tower. Time to reach the ground?

Goal: $t$ .
Strategy: Ground is $s=-20$ .

$-20=14.7t-4.9t^2$
$s=ut+\frac12at^2$ .
$4.9t^2-14.7t-20=0$
Rearrange.
$t=\frac{14.7+\sqrt{216.09+392}}{9.8}=4.02$
Positive root.

Answer

4.02

✓ CheckMore than the

3

s round trip ✓.

Worked exampleTime above 9.8 m

For the 14.7 m s $^{-1}$ projection, how long is the ball above 9.8 m?

Goal: Duration.
Strategy: Two times at that height.

$9.8=14.7t-4.9t^2\Rightarrow t^2-3t+2=0$
Divide by 4.9.
$t=1,\ 2$
Going up, coming down.

Answer

1

✓ CheckSymmetric about the top at

1.5

s ✓.

Common misconceptionMixed signs Choose a positive direction once and stick to it for

u

a

and

s

Pause and think

A stone falls from rest for 2 s. Distance?

Summary

$a=-9.8$ (up positive).
Top: $v=0$ .

Projected: Given an initial velocity.

HistoryThe Leaning Tower story. A famous story says Galileo dropped two balls of different mass from the Leaning Tower of Pisa to show they land together. It may never have happened, but in 1971 astronaut David Scott dropped a hammer and a feather on the Moon, with no air, and they did hit the ground together.

Fun factFalling faster than you think. After just

3

seconds of free fall (ignoring air resistance) an object is moving at about

29

m s⁻¹, over

100

km/h. Skydivers stop accelerating at around

55

m s⁻¹ because of air resistance.

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