Q3MechanicsCore

Vertical Motion Under Gravity

An object moving freely under gravity has constant acceleration $g$ downwards, so it is a suvat problem with the acceleration handed to you. What makes these questions error-prone is not the mathematics but the signs — and those are entirely under your control.

30 min Kinematics

The big picture

The single decision that determines whether a vertical-motion problem goes smoothly is choosing a positive direction and holding to it. Once that is fixed, the equations handle the entire flight in one go — rise, turning point and fall — without needing to be restarted at the top. Students who split the motion into two separate problems create twice the work and twice the opportunities for error, and often get the descent wrong when the object lands below its launch point. Trusting the sign convention is the whole skill.

What you'll be able to do

Model free vertical motion using $a = g$ downwards
Apply a consistent sign convention and interpret negative results
Find maximum height, time of flight and impact speed
Handle motion that finishes below the launch point

Setting up the sign convention

Choose which direction is positive before writing anything else, and state it. Both choices are equally valid; what matters is consistency.

Taking $upwards positive$ makes an upward launch velocity positive and the acceleration $a = - 9.8 m s^{- 2}$ , since gravity acts downwards. Taking $downwards positive$ makes $a = + 9.8$ and is convenient for objects that are simply dropped, since every quantity is then positive.

AQA questions normally use $g = 9.8 m s^{- 2}$ . Use whatever value the question specifies, keep it unrounded through the working, and round only at the end.

a = - g (upwards positive)

The usual choice for a launch.

a = + g (downwards positive)

Convenient for a drop.

Why this works

Objects in free flight all have the same acceleration regardless of mass — a heavy ball and a light one, released together, land together. That is why $g$ appears without any reference to the object’s mass.

Dropped objects

For an object released from rest, $u = 0$ and taking downwards positive keeps everything positive throughout. The equations simplify considerably.

The two standard results follow immediately: the distance fallen is $\frac{1}{2} g t^{2}$ , and the speed after falling a height $h$ is $2 g h$ .

s = \frac{1}{2} g t^{2}

Distance fallen from rest.

v = 2 g s

Speed after falling a height

s

1Downwards positive:

u = 0

a = 9.8

t = 2.5

s = \frac{1}{2} (9.8) (6.25) = 30.625

v = 0 + 9.8 (2.5) = 24.5 m s^{- 1}

AnswerHeight

30.6

m (3 s.f.); impact speed

24.5 m s^{- 1}

Tip — For a straightforward drop, take downwards as positive. Every quantity is then positive and there are no signs to lose.

Objects thrown upwards

At the highest point the velocity is momentarily $zero$ , but the acceleration is still $g$ downwards. Gravity does not pause; it is what brings the object back.

Setting $v = 0$ locates the top: the time to reach it is $\frac{u}{g}$ and the maximum height above the launch point is $\frac{u ^{2}}{2 g}$ .

If the object returns to its launch height, the motion is symmetric — the ascent and descent take equal times, and it arrives back at the same speed it left, but moving downwards.

t_{top} = \frac{u}{g}

Time to maximum height.

h_{max} = \frac{u ^{2}}{2 g}

Height gained above launch.

1Upwards positive:

u = 21

a = - 9.8

, and

v = 0

at the top.

h = \frac{2 1 ^{2}}{2 ( 9.8 )} = \frac{441}{19.6} = 22.5

3Time to the top:

\frac{21}{9.8} = 2.143

4By symmetry the total flight is twice that.

AnswerMaximum height

22.5

m; total time

4.29

s (3 s.f.).

Why this works

The symmetry is a consequence of constant acceleration, not a separate rule. The velocity changes at a steady rate, so it takes as long to fall from $0$ to $- 21$ as it took to rise from $21$ to $0$ .

Landing below the launch point

When an object is thrown from a cliff or a window, it does not return to its launch height and the symmetry is lost. The descent takes longer than the ascent, because the object continues past its starting level.

The reliable method is to handle the whole flight in one equation. With upwards positive, the final displacement is $negative$ — it finishes below where it started — and $s = u t + \frac{1}{2} a t^{2}$ becomes a quadratic in $t$ .

Splitting the flight into ascent and descent also works but takes longer and invites errors. Doubling the time to the top, however, is simply wrong here.

1Upwards positive:

u = 14.7

a = - 9.8

s = - 19.6

- 19.6 = 14.7 t - 4.9 t^{2}

, so

4.9 t^{2} - 14.7 t - 19.6 = 0

3Divide by 4.9:

t^{2} - 3 t - 4 = 0

, which factorises as

(t - 4) (t + 1) = 0

4Reject the negative root.

Answer

t = 4 s

Tip — A negative displacement is not an error — it is the sign convention telling you the object finished below its start. Forcing it positive is what produces the wrong answer.

Think like an examiner

State your sign convention in the first line of working.
Use the value of $g$ the question gives, and keep it unrounded until the final answer.
At maximum height set $v = 0$ , and say that the acceleration is still $g$ if asked.
Only use symmetry when the object returns to its launch height.

Common misconceptions

Using

a = + 9.8

with an upward positive velocity.

Gravity opposes upward motion, so

a = - 9.8

under that convention.

Saying acceleration is zero at the highest point.

The velocity is zero there; the acceleration is still

g

downwards.

Doubling the time to the top for an object landing below its launch point.

Symmetry needs equal start and finish heights. Otherwise solve the full quadratic.

Treating a negative displacement as impossible.

It means the object ended below the start, which is exactly what a cliff problem describes.

Vertical motion

a = \pm g

Sign set by your convention.

s = \frac{1}{2} g t^{2}

Dropped from rest.

v = 2 g s

Speed after falling

s

h_{m a x} = \frac{u ^{2}}{2 g}

Height gained from launch speed

u

t_{top} = \frac{u}{g}

Time to the highest point.

Stretch yourself

A ball is thrown vertically upwards at $19.6 m s^{- 1}$ . Find the two times at which it is $14.7 m$ above its launch point, and explain what the two answers represent. Take $g = 9.8 m s^{- 2}$ .

Hint — Use $s = u t + \frac{1}{2} a t^{2}$ with $s = 14.7$ and solve the quadratic. Expect two positive roots.

Questions students ask

Which direction should I take as positive?

Either, provided you state it and stay consistent. Downwards positive suits a simple drop; upwards positive suits a launch, since the initial velocity is then positive.

Does a heavier object fall faster?

No. With air resistance neglected, all objects accelerate at

g

regardless of mass, because the extra weight is exactly offset by the extra inertia.

Why is the acceleration not zero at the top?

Because gravity acts throughout the flight. The velocity passes through zero as it changes from upwards to downwards, but the rate of change is constant at

g

When can I use symmetry?

Only when the object finishes at its launch height. From a cliff or a window the descent is longer, and the full quadratic must be solved.

Key takeaways

Free vertical motion has constant acceleration $g$ downwards, independent of mass.
Choose and state a positive direction; the sign of $a$ follows from it.
At maximum height $v = 0$ but the acceleration is still $g$ .
Symmetry applies only when the object returns to its launch height.
Landing below the start gives a negative displacement and a quadratic in $t$ .

How this fits the course

Build on

Constant Acceleration

Variable Acceleration

Leads to

Variable Acceleration

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Vertical Motion Under Gravity

Setting up the sign convention

Dropped objects

Objects thrown upwards

Landing below the launch point

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Vertical Motion Under Gravity

Setting up the sign convention

Dropped objects

Objects thrown upwards

Landing below the launch point

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