Motion in a vertical plane with constant acceleration

Highlights only0 highlighted

This is an A Level section, and it opens the kinematics unit of Mechanics 2, the M2 paper (WME02).

In M1 every motion ran along one straight line.

Here a particle moves in a vertical plane: across and up or down at the same time, like a ball rolling off a table or a parcel dropped from a moving drone.

The acceleration is still constant, usually just gravity, but it now acts at an angle to the motion, so the path curves.

This section builds the one idea that makes all of that manageable, splitting the motion into two straight-line motions, and the next three sections use it for projectiles.

Kinematics in a plane, projectiles and variable acceleration together carried over a quarter of the marks in the June 2024, January 2025 and June 2025 M2 papers.

The idea: two straight-line motions at once

Watch a marble roll off the edge of a table $1.225$ m high at $3\ \text{m s}^{-1}$ .

Here is where it is every tenth of a second, measured across from the edge and down from the table top:

$t$ (s)

$0$

$0.1$

$0.2$

$0.3$

$0.4$

$0.5$

distance across (m)

$0$

$0.3$

$0.6$

$0.9$

$1.2$

$1.5$

distance fallen (m)

$0$

$0.049$

$0.196$

$0.441$

$0.784$

$1.225$

Read the two rows separately.

The distance across goes up by $0.3$ m every step: that is steady motion at $3\ \text{m s}^{-1}$ , exactly as if gravity did not exist.

The distance fallen is $4.9t^2$ : that is exactly how a marble dropped from rest falls, as if it were not moving across at all.

Put the two together and you get the curved path below.

Path of a marble that leaves the edge of a table 1.225 m high moving horizontally at 3 m s⁻¹. Dots mark its position every 0.1 s: they are 0.3 m apart horizontally every time, while the drops below the table top grow as 0.049, 0.196, 0.441, 0.784 and 1.225 m. It lands 1.5 m from the foot of the table after 0.5 s.

This is the whole of the method.

Gravity pulls straight down, so it changes the vertical velocity and does nothing to the horizontal velocity.

The two directions are independent: you can treat the horizontal motion and the vertical motion as two separate M1 problems that share one clock, the time $t$ .

The time is the only thing that links them.

A famous demonstration makes the same point.

Fire a ball horizontally from a table edge and, at the same instant, drop an identical ball from the same height.

They hit the floor together, however fast the first one was fired.

Its horizontal speed takes it further, but it does not keep it up for longer.

The model

The M2 paper models the moving object as a particle "moving freely under gravity".

That means:

the particle has no size, so it does not spin and air resistance is ignored;

the only force on it is its weight, so its acceleration is $g = 9.8\ \text{m s}^{-2}$ vertically downwards, a constant vector;

$g$ is the same everywhere along the path (the ground is flat and the heights are small).

Not every constant acceleration in a vertical plane is gravity alone.

If a steady horizontal wind pushes on a falling object with a constant force, the acceleration is still a constant vector, but now it has a horizontal part as well.

The syllabus wording, "constant acceleration, e.g. under gravity", covers both.

Gravity is the case you will meet most.

Vectors make it compact

Take a fixed origin $O$ , a unit vector $\mathbf{i}$ horizontal and $\mathbf{j}$ vertically upwards.

M2 questions that use vectors print a line in square brackets saying exactly this.

For a particle with initial velocity $\mathbf{u}$ and constant acceleration $\mathbf{a}$ , the M1 constant acceleration formulae hold as vector equations:

\mathbf{v} = \mathbf{u} + \mathbf{a}t \qquad\qquad \mathbf{r} = \mathbf{u}t + \tfrac{1}{2}\mathbf{a}t^2

where $\mathbf{r}$ is the displacement from the starting point.

Under gravity alone $\mathbf{a} = -9.8\mathbf{j}$ .

Writing $\mathbf{u} = u_x\mathbf{i} + u_y\mathbf{j}$ , the two equations split into components:

horizontal ( $\mathbf{i}$ )

vertical ( $\mathbf{j}$ , up positive)

acceleration

$0$

$-g$

velocity at time $t$

$u_x$ (constant)

$u_y - gt$

displacement at time $t$

$x = u_x t$

$y = u_y t - \tfrac{1}{2}gt^2$

Each column is ordinary straight-line motion.

The vertical column is the M1 vertical motion you already know; the horizontal column is constant speed.

All five suvat formulae may be used in the vertical column.

In the horizontal column only "distance = speed × time" is needed, because there is no acceleration.

One formula does not turn into a vector equation in the obvious way: $v^2 = u^2 + 2as$ .

Use it for one component at a time, for example $v_y^2 = u_y^2 + 2(-g)y$ for the vertical motion.

To get the speed, combine the components at the end with Pythagoras: speed $= \sqrt{v_x^2 + v_y^2}$ .

Why it works (not examined)

Acceleration is the rate of change of velocity.

If $\mathbf{a}$ is a constant vector, then $\dfrac{\mathrm{d}\mathbf{v}}{\mathrm{d}t} = \mathbf{a}$ integrates to $\mathbf{v} = \mathbf{a}t + \mathbf{c}$ , and $\mathbf{v} = \mathbf{u}$ at $t = 0$ gives $\mathbf{c} = \mathbf{u}$ .

Integrating $\dfrac{\mathrm{d}\mathbf{r}}{\mathrm{d}t} = \mathbf{u} + \mathbf{a}t$ again gives $\mathbf{r} = \mathbf{u}t + \tfrac{1}{2}\mathbf{a}t^2$ when $\mathbf{r} = \mathbf{0}$ at $t = 0$ .

A vector is integrated one component at a time, so each component obeys its own suvat.

Section 7 and section 8 of this unit do this integration for accelerations that change with time.

Why the path curves, and when it does not

The velocity changes by $-9.8\mathbf{j}$ every second.

If the particle starts with a horizontal part to its velocity, that part never changes while the vertical part grows downwards, so the direction of motion keeps turning towards the vertical: the path bends.

The shape is a parabola, because $x$ grows like $t$ while the drop grows like $t^2$ .

A particle released from rest with a constant acceleration does not curve at all.

With $\mathbf{u} = \mathbf{0}$ , $\mathbf{r} = \tfrac{1}{2}\mathbf{a}t^2$ is always a multiple of the fixed vector $\mathbf{a}$ , so it moves in a straight line in the direction of $\mathbf{a}$ .

A pebble dropped in a steady side wind falls along a sloping straight line, not a curve.

Worked example 3 uses this.

The marks, and the value of g

How the marks work

M1 is for a correct method applied to this question: a suvat equation with the right component of velocity in it. A1 is for an accurate equation or answer and needs its M1. B1 is an independent mark for a stated value or fact. DM1 (a capital D in M2 schemes) counts only if the earlier M1 it depends on was earned.

A1* marks a printed answer: every step to it must be shown and correct. A1ft or B1ft follows through from your own earlier wrong value. cso means correct solution only; oe means any equivalent form; isw means a correct answer followed by extra wrong working keeps its marks.

Some mark schemes use awrt (answers which round to); recent M2 schemes write "2 sf or 3 sf" instead. A misread that does not make the question easier costs two of the A or B marks in that part.

An equation earns its M1 only if it has the right terms, every term has the same dimensions, and every velocity or force that needs resolving has its sine or cosine. Many M2 schemes give a first A1 for an equation with at most one slip and a second A1 only when it is fully correct.

When a question names a method ("use the model", "use the work-energy principle"), another method scores nothing. Where the paper prints the warning that answers relying on calculator technology are not acceptable, every integral or derivative must be shown.

The g rule

The M2 cover gives $g = 9.8\ \text{m s}^{-2}$ and says an answer that used it goes to 2 or 3 significant figures. A fraction such as $\tfrac{10}{7}$ or a surd such as $\sqrt{392}$ left after using $9.8$ scores A0, and so does an over-long decimal.

When an answer is asked for "in terms of $g$ " (or $u$ , $a$ ), keep $g$ as a letter the whole way. Putting in $9.8$ halfway and writing $g$ back at the end is a common way to lose marks.

A question may tell you to model $g$ as $10\ \text{m s}^{-2}$ for that question only, with its own accuracy instruction. Then use $10$ in that question.

An answer that never used $g$ may be exact, or a decimal given to 3 s.f.

What the model leaves out

The model ignores air resistance, the spin of the ball, and the curvature of the Earth.

Air resistance acts against the velocity, so it has a horizontal part that slows the particle across as well as a vertical part.

Real balls therefore land short of the model's prediction, and the effect grows with speed.

When an M2 question asks you to criticise or refine the model, these are the points to make.

Where this leads

The same splitting into components drives every projectile question in this unit, the energy methods of the work and energy unit (which find speeds without needing the direction), and motion in a circle in M3.

In physics the independence of horizontal and vertical motion is the starting point for the motion of charged particles in electric fields, which follow exactly the same parabolas.

Worked example 4 is set out as an exam question; its diagram is Figure 1.

Figure 1

A ball is hit horizontally with speed U m s⁻¹ from point A, 2.45 m above horizontal ground. A net 0.9 m high stands 12 m from the point on the ground directly below A. A dashed curve shows the path of the ball passing over the net and landing beyond it.

Key terms

Vertical plane: a flat plane containing the vertical direction; a particle moving in it can move horizontally and vertically at the same time.

Moving freely under gravity: moving with no force acting except the particle's weight, so its acceleration is $g$ vertically downwards.

Components of velocity: the horizontal and vertical parts $u_x$ and $u_y$ of a velocity $\mathbf{u} = u_x\mathbf{i} + u_y\mathbf{j}$ ; the speed is $\sqrt{u_x^2 + u_y^2}$ .

Constant acceleration formulae (learn these, from M1): $v = u + at$ , $s = ut + \tfrac{1}{2}at^2$ , $s = vt - \tfrac{1}{2}at^2$ , $v^2 = u^2 + 2as$ , $s = \tfrac{1}{2}(u + v)t$ . They are on the M1 list of formulae students are expected to know, so they are not in the booklet.

Worked example 1: thrown horizontally from a cliff

A stone is thrown horizontally with speed $12\ \text{m s}^{-1}$ from a point at the top of a vertical cliff, $44.1$ m above the sea.

The stone is modelled as a particle moving freely under gravity.

Find

(a) the time taken for the stone to reach the sea, (2)
(b) the horizontal distance of the point where it lands from the foot of the cliff, (1)
(c) the speed of the stone as it reaches the sea. (3)

Downwards positive for the vertical motion; the initial vertical velocity is $0$ .

1. (a) Vertical: $s = ut + \tfrac{1}{2}at^2$ , so $44.1 = 0 + \tfrac{1}{2}(9.8)t^2$ M1

2. $t^2 = 9$ , $t = 3$ s A1

3. (b) Horizontal: $x = 12 \times 3 = 36$ m B1ft

4. (c) Vertical velocity: $v_y = 0 + 9.8 \times 3 = 29.4\ \text{m s}^{-1}$ B1

5. Speed $= \sqrt{12^2 + 29.4^2} = \sqrt{1008.36}$ M1

6. $= 31.75\ldots$ , so $32\ \text{m s}^{-1}$ (or $31.8\ \text{m s}^{-1}$ ) A1

Answer: $3$ s, $36$ m, $32\ \text{m s}^{-1}$ .

The time came only from the vertical motion: the $12\ \text{m s}^{-1}$ played no part in it.

The speed used $g$ , so it goes to 2 or 3 s.f.; $\sqrt{1008.36}$ would score A0.

The direction is $\tan^{-1}\dfrac{29.4}{12} = 68°$ below the horizontal: by the time it lands the stone is moving much more down than across.

Worked example 2: a velocity vector, and a printed time

[In this question $\mathbf{i}$ and $\mathbf{j}$ are horizontal and vertically upward unit vectors respectively.]

A particle $P$ is projected from a point $O$ with velocity $(6\mathbf{i} + 14.7\mathbf{j})\ \text{m s}^{-1}$ and moves freely under gravity.

(a) Find the position vector of $P$ relative to $O$ when $t = 2$ . (3)
(b) Show that $P$ is moving horizontally when $t = 1.5$ (2)
(c) Find the speed of $P$ when $t = 2$ (3)

1. (a) $\mathbf{r} = \mathbf{u}t + \tfrac{1}{2}\mathbf{a}t^2 = (6\mathbf{i} + 14.7\mathbf{j})t - 4.9t^2\mathbf{j}$ M1

2. $t = 2$ : $\mathbf{r} = 12\mathbf{i} + (29.4 - 19.6)\mathbf{j}$ A1

3. $\mathbf{r} = (12\mathbf{i} + 9.8\mathbf{j})$ m A1

4. (b) $\mathbf{v} = \mathbf{u} + \mathbf{a}t = 6\mathbf{i} + (14.7 - 9.8t)\mathbf{j}$ M1

5. When $t = 1.5$ the $\mathbf{j}$ component is $14.7 - 9.8(1.5) = 0$ , so the velocity is $6\mathbf{i}$ and $P$ is moving horizontally A1*

6. (c) $t = 2$ : $\mathbf{v} = 6\mathbf{i} + (14.7 - 19.6)\mathbf{j} = 6\mathbf{i} - 4.9\mathbf{j}$ M1

7. Speed $= \sqrt{6^2 + 4.9^2} = \sqrt{60.01}$ M1

8. $= 7.7\ \text{m s}^{-1}$ (or $7.75\ \text{m s}^{-1}$ ) A1

Answer: (a) $(12\mathbf{i} + 9.8\mathbf{j})$ m; (b) shown; (c) $7.7\ \text{m s}^{-1}$ .

The answer to (c) is a speed, a number, so leaving it as $6\mathbf{i} - 4.9\mathbf{j}$ loses the mark.

In (b) the last line says in words why a zero $\mathbf{j}$ component means "moving horizontally": that sentence is what the A1* is for.

At $t = 1.5$ the particle is at its highest point, $11.025$ m above $O$ .

Worked example 3: a side wind, released from rest

[In this question $\mathbf{i}$ and $\mathbf{j}$ are horizontal and vertically upward unit vectors respectively.]

A small parcel is released from rest from a drone hovering $19.6$ m above horizontal ground.

A steady wind acts on the parcel, and its acceleration is modelled as the constant vector $(1.4\mathbf{i} - 9.8\mathbf{j})\ \text{m s}^{-2}$ .

(a) Find the time taken for the parcel to reach the ground. (2)
(b) Find the horizontal distance between the point where the parcel lands and the point directly below the drone. (2)
(c) Show that the parcel moves in a straight line inclined at $\tan^{-1}\tfrac{1}{7}$ to the vertical. (2)

1. (a) Vertical, downwards positive: $19.6 = \tfrac{1}{2}(9.8)t^2$ M1

2. $t^2 = 4$ , $t = 2$ s A1

3. (b) Horizontal: $x = \tfrac{1}{2}(1.4)t^2 = 0.7(2^2)$ M1

4. $x = 2.8$ m A1

5. (c) With $\mathbf{u} = \mathbf{0}$ , $\mathbf{r} = \tfrac{1}{2}(1.4\mathbf{i} - 9.8\mathbf{j})t^2 = 0.7t^2(\mathbf{i} - 7\mathbf{j})$ M1

6. Every position is a multiple of the fixed vector $\mathbf{i} - 7\mathbf{j}$ , so the path is a straight line, and it makes an angle $\tan^{-1}\tfrac{1}{7}$ with the vertical A1*

Answer: (a) $2$ s; (b) $2.8$ m; (c) shown.

Both answers used $g$ , but $2$ and $2.8$ already have 2 significant figures, so they stand as they are: $t^2 = \tfrac{19.6}{4.9} = 4$ exactly.

The fall time is the same as with no wind, because the wind acts horizontally: it moves the landing point, not the landing time.

The ratio $1.4 : 9.8 = 1 : 7$ of the accelerations is the slope of the path, measured from the vertical.

Worked example 4: clearing a net

A ball is hit horizontally from a point $A$ , which is $2.45$ m above horizontal ground, as shown in Figure 1.

A vertical net of height $0.9$ m stands $12$ m from the point on the ground directly below $A$ .

The ball is hit towards the net with speed $U\ \text{m s}^{-1}$ and moves freely under gravity in the vertical plane at right angles to the net.

(a) Find the time taken for the ball to fall from $A$ to the ground. (2)
(b) Find the least value of $U$ for which the ball passes over the net. (4)

For this least value of $U$ ,

1. (a) Vertical, downwards positive: $2.45 = \tfrac{1}{2}(9.8)t^2$ M1

2. $t^2 = 0.5$ , $t = 0.707\ldots$ , so $0.71$ s A1

3. (b) To pass over the net the ball may fall at most $2.45 - 0.9 = 1.55$ m by the time it reaches the net B1

4. Vertical: $1.55 = 4.9T^2$ , so $T = 0.5624\ldots$ s, where $T$ is the time to reach the net M1

5. Horizontal: $12 = UT$ DM1

6. $U = \dfrac{12}{0.5624\ldots} = 21.3\ldots$ , so $U = 21$ (or $21.3$ ) A1

7. (c) Horizontal distance to landing $= 21.336\ldots \times 0.7071\ldots$ M1

8. $= 15.08\ldots$ m A1ft

9. Beyond the net: $15.08\ldots - 12 = 3.1$ m (or $3.09$ m) A1

Answer: (a) $0.71$ s; (b) $21\ \text{m s}^{-1}$ ; (c) $3.1$ m.

In (b) the net condition was turned into a statement about the vertical drop first, then the horizontal motion gave $U$ .

A smaller $U$ means the ball takes longer to reach the net and so drops further.

Check with $U = 20$ : it reaches the net after $0.6$ s, having fallen $1.764$ m, so it is only $0.686$ m up and hits the net.

Keep the unrounded $U$ and $t$ in (c); rounding $U$ to $21$ first gives $2.8$ m, which is wrong.

Exam tip

Write two columns, "horizontal" and "vertical", before any numbers, and fill in $u$ , $a$ and $s$ for each.

Examiners give the M1 for a suvat line that uses the vertical component with $g$ , or the horizontal component with no acceleration; a vertical equation with the horizontal velocity in it does not earn that mark.

Then link the columns only through $t$ .

Common mistakes

Putting the horizontal speed into the vertical equation. The speed the stone is thrown at feels like "the" $u$ , so it gets used everywhere. Fix: a particle thrown horizontally has vertical $u = 0$ ; the throw speed belongs only in the horizontal column.

Thinking a faster throw stays up longer. It travels further, so it seems to be in the air longer. Fix: the fall time depends only on the vertical motion; horizontal speed changes where it lands, not when.

Adding the components to get the speed. $12 + 29.4$ is not the speed. Fix: speed $= \sqrt{v_x^2 + v_y^2}$ , and a speed is a number, never a vector.

Leaving $\sqrt{1008.36}$ or $\tfrac{10}{7}$ as a final answer after using $g = 9.8$ . Fix: round to 2 or 3 s.f.; only answers that never used $g$ may stay exact.

Rounding an intermediate time, then using it. Fix: keep full calculator values to the last line, as in Worked example 4(c).

Check your understanding

1. A stone is thrown horizontally with speed $8\ \text{m s}^{-1}$ from the top of a vertical cliff. It moves freely under gravity and hits the sea $2.5$ s later.
(a) Find the height of the cliff above the sea. (2)
(b) Find the horizontal distance of the point where the stone hits the sea from the foot of the cliff. (1)
(c) Find the speed of the stone as it hits the sea. (3) Total [6]

2. [In this question $\mathbf{i}$ and $\mathbf{j}$ are horizontal and vertically upward unit vectors respectively.] A particle is projected from a point $O$ with velocity $(5\mathbf{i} + 9.8\mathbf{j})\ \text{m s}^{-1}$ and moves freely under gravity.
(a) Show that the particle is level with $O$ again when $t = 2$ (3)
(b) Find the speed of the particle when $t = 2$ (3) Total [6]

3. [In this question $\mathbf{i}$ and $\mathbf{j}$ are horizontal and vertically upward unit vectors respectively.] A small ball is thrown with velocity $4.2\mathbf{i}\ \text{m s}^{-1}$ from a point $10$ m above horizontal ground. A steady wind blows against it, and its acceleration is modelled as the constant vector $(-0.7\mathbf{i} - 9.8\mathbf{j})\ \text{m s}^{-2}$ .
(a) Find the time taken for the ball to reach the ground. (2)
(b) Find the horizontal distance travelled by the ball before it reaches the ground. (3)
(c) Find the speed of the ball as it reaches the ground. (3) Total [8]

4. A particle is projected horizontally with speed $14\ \text{m s}^{-1}$ from a point at the top of a vertical tower of height $h$ metres, standing on horizontal ground. The particle moves freely under gravity and hits the ground moving at $45°$ to the horizontal.
(a) Show that $h = 10$ (4)
(b) Find the distance from the foot of the tower to the point where the particle hits the ground. (2) Total [6]

5. Understanding check. Two identical balls are held at the same height, $4.9$ m above horizontal ground. At the same instant, ball $A$ is released from rest and ball $B$ is thrown horizontally at $10\ \text{m s}^{-1}$ . Both are modelled as particles moving freely under gravity. A student says that $B$ lands later than $A$ because it travels further.
(a) Show that the student is wrong. (3)
(b) Find the distance between the points where $A$ and $B$ land. (1)
(c) In reality air resistance acts on the balls. By considering the direction of the air resistance on $B$ , show that $B$ lands short of the point the model predicts. (2) Total [6]

6. A particle is released from rest and moves with constant acceleration $(2.1\mathbf{i} - 9.8\mathbf{j})\ \text{m s}^{-2}$ , where $\mathbf{i}$ and $\mathbf{j}$ are horizontal and vertically upward unit vectors respectively.
(a) Find the size of the angle between its path and the downward vertical. (2)
(b) Find the speed of the particle $3$ s after it is released. (2)
(c) Find the distance of the particle from its point of release $3$ s after it is released. (2) Total [6]

Answers

1. (a) Vertical, downwards positive: $s = 0 + \tfrac{1}{2}(9.8)(2.5)^2$ M1 $= 30.625$ , so $31$ m (or $30.6$ m) A1. (b) $8 \times 2.5 = 20$ m B1. (c) $v_y = 9.8 \times 2.5 = 24.5$ B1, speed $= \sqrt{8^2 + 24.5^2}$ M1 $= 26\ \text{m s}^{-1}$ (or $25.8$ ) A1. The height used $g$ , so $30.625$ scores A0: cut it to 2 or 3 s.f.

2. (a) Vertical: $y = 9.8t - 4.9t^2$ M1 A1; when $t = 2$ , $y = 19.6 - 19.6 = 0$ , so the particle is level with $O$ A1*. (b) $\mathbf{v} = 5\mathbf{i} + (9.8 - 9.8 \times 2)\mathbf{j} = 5\mathbf{i} - 9.8\mathbf{j}$ M1, speed $= \sqrt{5^2 + 9.8^2}$ M1 $= 11\ \text{m s}^{-1}$ (or $11.0$ ) A1. It is back at its starting speed, moving down as steeply as it went up.

3. (a) Vertical: $10 = 4.9t^2$ M1, $t = 1.428\ldots$ , so $1.4$ s (or $1.43$ s) A1. (b) $x = 4.2t - 0.35t^2$ M1 A1 $= 6 - 0.714\ldots = 5.3$ m (or $5.29$ m) A1. (c) $v_x = 4.2 - 0.7(1.428\ldots) = 3.2$ and $v_y = -9.8(1.428\ldots) = -14$ M1, speed $= \sqrt{3.2^2 + 14^2}$ DM1 $= 14\ \text{m s}^{-1}$ (or $14.4$ ) A1.

4. (a) At $45°$ the vertical speed equals the horizontal speed, $14\ \text{m s}^{-1}$ B1. Vertical: $14^2 = 0 + 2(9.8)h$ M1 A1, $h = \dfrac{196}{19.6} = 10$ A1*. (b) Time: $14 = 9.8t$ , $t = 1.428\ldots$ M1; distance $= 14t = 20$ m A1.

5. (a) Both balls start with vertical velocity $0$ and have acceleration $g$ downwards B1; for each, $4.9 = \tfrac{1}{2}(9.8)t^2$ , so $t = 1$ s M1; the times are equal, so the balls land together and the student is wrong: $B$ 's horizontal velocity changes where it lands, not when A1*. (b) $A$ lands directly below its start and $B$ lands $10 \times 1 = 10$ m from that point B1. (c) Air resistance acts against the velocity, so on $B$ it has a horizontal part opposing the motion across B1; $B$ 's horizontal velocity falls below $10\ \text{m s}^{-1}$ during the flight, so it covers less than the model's $10$ m B1.

6. (a) From rest the path is along $\mathbf{a}$ , so the angle with the downward vertical is $\tan^{-1}\dfrac{2.1}{9.8}$ M1 $= 12°$ (or $12.1°$ ) A1. (b) $\mathbf{v} = 3(2.1\mathbf{i} - 9.8\mathbf{j}) = 6.3\mathbf{i} - 29.4\mathbf{j}$ M1, speed $= \sqrt{6.3^2 + 29.4^2} = 30\ \text{m s}^{-1}$ (or $30.1$ ) A1. (c) $\mathbf{r} = \tfrac{1}{2}(2.1\mathbf{i} - 9.8\mathbf{j})(3^2) = 9.45\mathbf{i} - 44.1\mathbf{j}$ M1, distance $= \sqrt{9.45^2 + 44.1^2} = 45$ m (or $45.1$ m) A1.

Motion in a vertical plane with constant acceleration

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