What is sign confusion on the vertical displacement?

Decide once whether up or down is positive and apply it consistently. For a launch and landing at the same level, the net vertical displacement is zero, not the maximum height.

A ball is kicked at 14 m s^-1 at 45^. Find its range on level ground (g = 9.8). [3 marks]

A projectile launched at 25^ has initial vertical velocity 10 m s^-1. Find its greatest height (g = 9.8). [2 marks]

SQA AH style: range and height-style practice: A ball is projected from ground level at 20 m s^-1 at 30^ above the horizontal. Taking g = 9.8 m s^-2, find the time of flight, the horizontal range, and the greatest height reached.

Resolve: u_x = 20cos 30^ = 10sqrt(3), u_y = 20sin 30^ = 10 m s^-1 (1 mark). Time of flight: vertical displacement returns to zero, 0 = u_y t - 12gt^2, so t = 2u_yg = 209.8 ≈ 2.04 s (2 marks). Range: R = u_x t = 10sqrt(3)× 2.04 ≈ 35.3 m (1 mark). Greatest height: at the top v_y = 0, so v_y^2 = u_y^2 - 2gH gives H = u_y^22g = 10019.6 ≈ 5.10 m (2 marks). Markers reward resolving the velocity, treating the two directions independently, and using v_y = 0 at the highest point.

ScotlandMathematics of MechanicsSyllabus dot point

How do you model a projectile moving under gravity, and how do you find its range, maximum height, time of flight and path?

Model projectile motion under gravity by treating horizontal and vertical motion independently; find time of flight, range, maximum height, velocity at any time, and the equation of the parabolic path.

A focused answer to the SQA Advanced Higher Mathematics of Mechanics content on projectiles, resolving the initial velocity into horizontal and vertical parts, applying the constant-acceleration equations independently in each direction, and deriving time of flight, range, maximum height and the equation of the trajectory.

Generated by Claude Opus 4.813 min answerUpdated 2026-06-16

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Resolving and separating the motion
Maximum height, time of flight and range
The equation of the path
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What this dot point is asking

A projectile is anything moving freely under gravity once it has been launched, with air resistance ignored. The SQA wants you to model it by splitting the motion into two independent parts: horizontal motion at constant velocity, and vertical motion with constant downward acceleration $g$ . From that split come all the standard results - time of flight, range, maximum height, the velocity at any instant, and the parabolic equation of the path.

Resolving and separating the motion

The key modelling idea is that gravity acts only downward, so it changes only the vertical velocity. The horizontal velocity stays constant throughout the flight.

Because the two sets of equations share only the time $t$ , the time is the bridge between them: solve a vertical condition for $t$ , then feed that $t$ into the horizontal equation (or vice versa).

Stone thrown horizontally from a cliff

A stone is thrown horizontally at $15$ m s $^{-1}$ from a cliff $20$ m high. Find how far from the base it lands ( $g = 9.8$ ).

Step 1: Set up the two independent directions

Gravity acts only downward, so it changes only the vertical component. The horizontal velocity stays constant throughout. Take upward as positive:

\text{Horizontal: } u_x = 15,\ x = 15t. \quad \text{Vertical: } u_y = 0,\ y = -\tfrac{1}{2}(9.8)t^2.

Step 2: Use the vertical drop to find the flight time

The stone must fall $20\ \text{m}$ , so set the vertical displacement to $-20$ :

-20 = -4.9t^2, \quad t^2 = \frac{20}{4.9}, \quad t \approx 2.02\ \text{s.}

Step 3: Find the horizontal range

The horizontal distance is simply the constant horizontal speed multiplied by the time found in step 2:

x = 15 \times 2.02 \approx 30.3\ \text{m from the base of the cliff.}

Final answer: the stone lands approximately $30.3\ \text{m}$ from the base of the cliff.

Maximum height, time of flight and range

These standard results all come from applying the right condition to the vertical motion.

The maximum height comes from $v_y^2 = u_y^2 - 2gH$ with $v_y = 0$ , giving $H = \dfrac{u_y^2}{2g}$ . The time of flight on level ground comes from setting the vertical displacement back to zero, $t = \dfrac{2u_y}{g}$ . The range is then the horizontal distance covered in that time, $R = u_x t = \dfrac{2u_x u_y}{g}$ , which (using $2\sin\theta\cos\theta = \sin 2\theta$ ) equals $\dfrac{V^2\sin 2\theta}{g}$ and is greatest at $\theta = 45^\circ$ .

The equation of the path

Eliminating the time between the horizontal and vertical equations gives $y$ as a function of $x$ . The result is a downward parabola, which is why projectile motion is described as parabolic.

y = x\tan\theta - \frac{gx^2}{2V^2\cos^2\theta}.

This form is useful for "does it clear the wall" problems: substitute the wall's horizontal distance for $x$ and compare the resulting $y$ with the wall's height.

Try this

Q1. A ball is kicked at $14$ m s $^{-1}$ at $45^\circ$ . Find its range on level ground ( $g = 9.8$ ). [3 marks]

Cue. $R = \dfrac{V^2\sin 2\theta}{g} = \dfrac{196\sin 90^\circ}{9.8} = 20$ m.

Q2. A projectile launched at $25^\circ$ has initial vertical velocity $10$ m s $^{-1}$ . Find its greatest height ( $g = 9.8$ ). [2 marks]

Cue. $H = \dfrac{u_y^2}{2g} = \dfrac{100}{19.6} \approx 5.1$ m.

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AH style: range and height6 marksA ball is projected from ground level at

20

m s

^{-1}

30^\circ

above the horizontal. Taking

g = 9.8

m s

^{-2}

, find the time of flight, the horizontal range, and the greatest height reached.

Show worked answer →

Resolve: $u_x = 20\cos 30^\circ = 10\sqrt{3}$ , $u_y = 20\sin 30^\circ = 10$ m s $^{-1}$ (1 mark).

Time of flight: vertical displacement returns to zero, $0 = u_y t - \tfrac{1}{2}gt^2$ , so $t = \dfrac{2u_y}{g} = \dfrac{20}{9.8} \approx 2.04$ s (2 marks).

Range: $R = u_x t = 10\sqrt{3}\times 2.04 \approx 35.3$ m (1 mark).

Greatest height: at the top $v_y = 0$ , so $v_y^2 = u_y^2 - 2gH$ gives $H = \dfrac{u_y^2}{2g} = \dfrac{100}{19.6} \approx 5.10$ m (2 marks). Markers reward resolving the velocity, treating the two directions independently, and using $v_y = 0$ at the highest point.

AH style: trajectory equation5 marksA projectile is launched from the origin with speed

V

at angle

\theta

. Show that its path has equation

y = x\tan\theta - \dfrac{gx^2}{2V^2\cos^2\theta}

Show worked answer →

Horizontal: $x = V\cos\theta\, t$ , so $t = \dfrac{x}{V\cos\theta}$ (1 mark).

Vertical: $y = V\sin\theta\, t - \tfrac{1}{2}gt^2$ (1 mark).

Substitute for $t$ : $y = V\sin\theta\cdot\dfrac{x}{V\cos\theta} - \tfrac{1}{2}g\left(\dfrac{x}{V\cos\theta}\right)^2$ (2 marks).

Simplify: the first term is $x\tan\theta$ and the second is $\dfrac{gx^2}{2V^2\cos^2\theta}$ , giving $y = x\tan\theta - \dfrac{gx^2}{2V^2\cos^2\theta}$ (1 mark). Markers reward eliminating $t$ between the two component equations and the correct simplification, recognising the path is a parabola in $x$ .

Related dot points

Sources & how we know this

Advanced Higher Mathematics of Mechanics Course Specification (C802 77) — SQA (2019)