What is a projectile in the context of Edexcel International A Levels Mechanics 2?

In Edexcel International A Levels Mechanics 2, a projectile is an object that is thrown or projected into the air and is subject to only the force of gravity and air resistance. The study of projectiles involves analyzing their motion, which is typically parabolic, and understanding how initial velocity, angle of projection, and gravitational acceleration affect their trajectory.

How do you calculate the range of a projectile in Mechanics 2?

The range of a projectile in Mechanics 2 can be calculated using the formula R = (v^2 * sin(2θ)) / g, where v is the initial velocity, θ is the angle of projection, and g is the acceleration due to gravity. This formula assumes that the projectile is launched from and lands on the same horizontal level, and air resistance is negligible.

What factors affect the time of flight of a projectile?

The time of flight of a projectile is affected by the initial velocity, the angle of projection, and the acceleration due to gravity. It can be calculated using the formula T = (2v * sin(θ)) / g, where v is the initial velocity, θ is the angle of projection, and g is the gravitational acceleration. The time of flight is longer for higher initial velocities and angles closer to 45 degrees.

Why is the angle of 45 degrees often considered optimal for maximum range in projectile motion?

In projectile motion, the angle of 45 degrees is considered optimal for maximum range because it provides the best balance between vertical and horizontal components of the initial velocity. At this angle, the projectile achieves the maximum horizontal distance, assuming no air resistance and that it is launched from and lands on the same level.

How does air resistance affect the motion of a projectile?

Air resistance affects the motion of a projectile by reducing its range, altering its trajectory, and decreasing its maximum height and time of flight. Unlike ideal projectile motion, where only gravity acts on the projectile, air resistance causes the projectile to slow down, resulting in a path that is less than the theoretical parabolic trajectory. In Edexcel International A Levels Mechanics 2, air resistance is often neglected for simplicity unless specified otherwise.

Why is "Using $a = +g$ for vertical motion when upward is taken positive" a common mistake in Projectiles?

Why it happens: Students see 9.8 as 'the acceleration' and forget gravity acts downward. How to avoid it: State your sign convention at the top of the working: 'taking upward as positive, a = -9.8\,m s^-2'. Mark schemes credit clear conventions and apply ECF if the convention is consistent. Source: WME02/01 examiner reports — flagged annually

Why is "Treating the speed $u$ as both horizontal and vertical components" a common mistake in Projectiles?

Why it happens: Forgetting that velocity is a vector and must be decomposed. How to avoid it: First line of every projectile solution: u_x = u, u_y = u. Draw the right-angled triangle of components.

Why is "Writing the trajectory equation as $y = x\tan\theta - \dfrac{gx^{2}}{2u^{2}}$ (missing the $\cos^{2}\theta$)" a common mistake in Projectiles?

Why it happens: Memorisation slip — students confuse it with the simpler vertical-motion formula. How to avoid it: Re-derive: from x = u · t, t = x/(u). Substitute into y = u · t - 12g t^2. The ^2 comes from t^2. Source: WME02/01 examiner reports — recurring

Why is "Using horizontal/vertical axes for projection on an inclined plane" a common mistake in Projectiles?

Why it happens: Default habit from level-ground problems. How to avoid it: On an inclined plane, ALWAYS use axes along and perpendicular to the plane. The 'landing condition' is then simply perpendicular displacement = 0, exactly as on level ground.

Why is "Taking the vertical-velocity-at-impact as the impact speed" a common mistake in Projectiles?

Why it happens: Forgetting the horizontal component is still there. How to avoid it: Impact speed = √(v_x^2) + v_y^2. v_x is constant throughout the flight (no air resistance); v_y changes. Always combine.

Mechanics 2Edexcel International A Levels Mathematics (XMA01-YMA01)

Projectiles

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Projectiles Study Notes — Edexcel IAL Mathematics M2 (WME02/01, 2018 spec — 2026 onwards)

Model a projectile as independent horizontal (constant velocity) and vertical (acceleration $g$ downward) motion. Apply suvat in each direction. Topics: time of flight, range, greatest height, trajectory equation, projection from a height, projection on an inclined plane. Air resistance neglected throughout.

At a glance

Split into components: $u_x = u\cos\theta$ (constant), $u_y = u\sin\theta$ (changes with gravity).
Vertical motion: $a = -g$ (upward positive). Horizontal motion: $a = 0$ .
Time of flight (level): $T = \dfrac{2u\sin\theta}{g}$ . Range (level): $R = \dfrac{u^{2}\sin 2\theta}{g}$ .
Greatest height: $H = \dfrac{u^{2}\sin^{2}\theta}{2g}$ , reached when $v_y = 0$ .
Trajectory equation: $y = x\tan\theta - \dfrac{gx^{2}}{2u^{2}\cos^{2}\theta}$ .
From a height: use suvat vertically with $s = -h$ (or $+h$ downward).
Inclined plane: rotate axes along/perpendicular to the slope.
Air resistance: ALWAYS neglected at M2.

What you’ll learn

Mapped to the Cambridge IGCSE XMA01-YMA01 syllabus (2026 onwards).

M2 1.1 — Model the motion of a projectile by treating horizontal and vertical motion separately, ignoring air resistance.
M2 1.2 — Find the time of flight, horizontal range and greatest height of a projectile launched from a horizontal plane.
M2 1.3 — Derive and use the equation of the trajectory of a projectile.
M2 1.4 — Solve problems involving projection from a height and projection on an inclined plane.

Setting up: resolving the initial velocity

Components first. Then use suvat in each direction independently.

The model. A projectile is a particle that, after launch, is acted on only by gravity — no air resistance, no other forces. Its acceleration is therefore constant: $g \approx 9.8\,\text{m s}^{-2}$ vertically downward, $0$ horizontally.

Components. Launch speed $u$ at angle $\theta$ above the horizontal:

$u_x = u\cos\theta, \qquad u_y = u\sin\theta.$

Always draw the right-angled triangle:

          u
        ┌───
       /│
      / │ u sin θ
     /  │
    /θ  │
   └────┘
   u cos θ

Sign convention. Upward positive is conventional. State your convention explicitly in the working — Edexcel mark schemes reward clarity.

Time-independence. Horizontal and vertical motions are completely independent. The same time $t$ links them: $x = u_x t$ and $y = u_y t - \tfrac{1}{2}g t^{2}$ .

Why air resistance is neglected. At GCSE the assumption is implicit; at M2 it's stated. Real projectiles experience drag $\propto v^{2}$ or $\propto v$ (Stokes), but this is deliberately excluded from the IAL specification. Any answer that "accounts for air resistance" loses marks because it doesn't match the model.

Components: $u\cos\theta$ horizontal, $u\sin\theta$ vertical.
Horizontal velocity is CONSTANT throughout the flight.
Vertical velocity changes by $-g$ per second.
Declare your sign convention.

Projection from level ground

Time of flight, range, greatest height — three standard formulae or three short suvat applications.

Time of flight $T$ . Particle returns to $y = 0$ . From $y = u\sin\theta \cdot t - \tfrac{1}{2}g t^{2} = 0$ , factor out $t$ :

$t\left(u\sin\theta - \tfrac{1}{2}g t\right) = 0 \Rightarrow T = \dfrac{2u\sin\theta}{g}.$

Range $R$ . Horizontal distance covered in time $T$ :

$R = u\cos\theta \cdot T = \dfrac{2u^{2}\sin\theta\cos\theta}{g} = \dfrac{u^{2}\sin 2\theta}{g}.$

Maximised at $\theta = 45^{\circ}$ (since $\sin 90^{\circ} = 1$ ).
Two angles $\theta$ and $90^{\circ} - \theta$ give the same range — useful when a question asks "find the two possible angles".

Greatest height $H$ . At peak, $v_y = 0$ . Use $v^{2} = u^{2} + 2as$ vertically:

$0 = (u\sin\theta)^{2} - 2gH \Rightarrow H = \dfrac{u^{2}\sin^{2}\theta}{2g}.$

Alternative: $H = \tfrac{1}{2}g(T/2)^{2}$ from symmetry.

Worked example. $u = 28$ m/s, $\theta = 30^{\circ}$ , $g = 9.8$ .

Components: $u_x = 14\sqrt{3}$ , $u_y = 14$ .
$T = 28/9.8 = 20/7 \approx 2.86$ s.
$R = u_x T = 14\sqrt{3} \cdot 20/7 = 40\sqrt{3} \approx 69.3$ m.
$H = 14^{2}/(2 \cdot 9.8) = 196/19.6 = 10$ m.

Time of flight: solve $y = 0$ vertically.
Range: $R = u_x T$ — never $R = uT$ .
Greatest height: $v_y = 0$ at peak.
Range max at $45^{\circ}$ ; two angles give same range.

The trajectory equation

Eliminate $t$ to get $y$ in terms of $x$ — useful when the path is constrained.

Derivation. From $x = u\cos\theta \cdot t$ , isolate $t$ :

$t = \dfrac{x}{u\cos\theta}.$

Substitute into $y = u\sin\theta \cdot t - \tfrac{1}{2}g t^{2}$ :

$y = u\sin\theta \cdot \dfrac{x}{u\cos\theta} - \dfrac{g x^{2}}{2 u^{2} \cos^{2}\theta} = x\tan\theta - \dfrac{g x^{2}}{2 u^{2} \cos^{2}\theta}.$

This is the equation of the trajectory — a parabola opening downward.

Using $\sec^{2}\theta = 1 + \tan^{2}\theta$ . When a question gives a point $(x_0, y_0)$ on the path and asks for $\theta$ , rewrite:

$y = x\tan\theta - \dfrac{g x^{2}}{2 u^{2}}(1 + \tan^{2}\theta).$

Now you have a quadratic in $T = \tan\theta$ . Two solutions → two angles of projection that reach the same point.

Worked example. $u = 30$ , passes through $(20, 5)$ , $g = 9.8$ :

$5 = 20T - \dfrac{9.8 \cdot 400}{1800}(1 + T^{2}) \Rightarrow 98 T^{2} - 900 T + 323 = 0.$

Solving: $T \approx 8.81$ or $T \approx 0.374$ , giving $\theta \approx 83.5^{\circ}$ or $\theta \approx 20.5^{\circ}$ .

Geometric meaning. The two roots are the high and low trajectories. Halfway between them (in angle, sort of) is the maximum-range parabola — the envelope of all parabolas reachable from the launch point with the given speed.

Trajectory: $y = x\tan\theta - \dfrac{gx^{2}}{2u^{2}\cos^{2}\theta}$ .
Use $\sec^{2} = 1 + \tan^{2}$ for find-the-angle problems.
Two roots = high and low trajectory.
Path is always a parabola (no air resistance).

Projection from a height (cliff / building)

Use suvat vertically with non-zero final displacement.

Setup. Take origin at the launch point, $y$ upward. The landing point is at $y = -h$ (where $h$ > 0 is the height of the cliff).

Time of flight. From $y = u_y t - \tfrac{1}{2}g t^{2} = -h$ :

$\tfrac{1}{2}g t^{2} - u_y t - h = 0 \Rightarrow t = \dfrac{u_y + \sqrt{u_y^{2} + 2gh}}{g}.$

Take the positive root (the negative one corresponds to extrapolating motion backward in time).

Horizontal range. $R = u_x t$ as usual.

Speed at impact. Two methods:

Components: $v_x = u_x$ , $v_y = u_y - gt$ . Impact speed $= \sqrt{v_x^{2} + v_y^{2}}$ .
Energy conservation (cross-check): $\tfrac{1}{2}v^{2} = \tfrac{1}{2}u^{2} + gh$ , so $v = \sqrt{u^{2} + 2gh}$ .

Note that energy conservation gives the impact speed directly — independent of the launch angle. A surprising and useful fact for marking schemes that say "show that the impact speed is ...".

Worked example. Stone thrown from $40$ m cliff at $15$ m/s at $\alpha$ below horizontal, $\sin\alpha = 3/5$ .

Components (with $y$ up): $u_x = 12$ , $u_y = -9$ .
Vertical: $-40 = -9t - 4.9 t^{2}$ , i.e. $4.9 t^{2} + 9t - 40 = 0$ .
$t = \dfrac{-9 + \sqrt{81 + 784}}{9.8} = \dfrac{-9 + \sqrt{865}}{9.8} \approx 2.08$ s.
Horizontal range: $12 \cdot 2.08 \approx 25.0$ m.
Impact speed: $v = \sqrt{15^{2} + 2 \cdot 9.8 \cdot 40} = \sqrt{225 + 784} = \sqrt{1009} \approx 31.8$ m/s.

Origin at launch; landing at $y = -h$ .
Solve the resulting quadratic in $t$ ; take positive root.
Impact speed = $\sqrt{u^{2} + 2gh}$ by energy conservation.
Cross-check using both methods on full-mark questions.

Projection on an inclined plane

Rotate axes: along and perpendicular to the plane. The 'landing condition' simplifies.

Standard trick. When a projectile is launched on a slope, rotate your coordinate axes so that one axis runs along the slope and the other is perpendicular to it.

Gravity components. If the slope makes angle $\alpha$ with the horizontal, gravity has components:

Along the slope (down-slope positive): $g\sin\alpha$ .
Perpendicular to slope (into slope positive): $g\cos\alpha$ .

Velocity components. If the particle is projected at angle $\beta$ above the slope (so $\alpha + \beta$ above horizontal), with speed $u$ :

Along slope: $u\cos\beta$ .
Perpendicular to slope: $u\sin\beta$ .

Landing condition. Particle lands when perpendicular displacement returns to zero — exactly as level-ground time-of-flight:

$0 = u\sin\beta \cdot t - \tfrac{1}{2}(g\cos\alpha) t^{2} \Rightarrow t = \dfrac{2u\sin\beta}{g\cos\alpha}.$

Distance up the plane.

$s = u\cos\beta \cdot t - \tfrac{1}{2}(g\sin\alpha) t^{2}.$

(Note the minus: the along-slope component of gravity slows the particle as it travels up the plane.)

Worked example. Plane at $\alpha = 20^{\circ}$ ; particle projected at $u = 25$ m/s, $\beta = 40^{\circ}$ above plane, $g = 9.8$ .

$t = \dfrac{2 \cdot 25 \sin 40^{\circ}}{9.8 \cos 20^{\circ}} \approx \dfrac{32.14}{9.21} \approx 3.49$ s.
$s = 25\cos 40^{\circ} \cdot 3.49 - \tfrac{1}{2} \cdot 9.8\sin 20^{\circ} \cdot 3.49^{2} \approx 66.85 - 20.43 \approx 46.4$ m.

Rotate axes along/perpendicular to plane.
$g \to (g\sin\alpha, g\cos\alpha)$ in those directions.
Landing: perpendicular displacement $= 0$ .
Use $\beta$ = angle above the plane (NOT above horizontal).

How it’s examined

Projectiles is the anchor question of every WME02/01 M2 paper, usually Q4 or Q6 and worth $8$ - $14$ marks. The level-ground formulae alone earn ~6 marks; the harder parts test projection from a height (with a quadratic in $t$ ) or projection on an inclined plane. Examiners reward a clear diagram with components labelled — many candidates lose M-marks because they don't show how they decomposed the initial velocity. ECF is generous: an early arithmetic slip carries through and only one A-mark is typically lost. A* candidates should target full marks: this is the most predictable topic on the paper.

Sources: Pearson Edexcel International Advanced Level Mathematics specification (2018 — current for 2026 onwards); Edexcel IAL Mathematical Formulae and Statistical Tables (2018); WME02/01 Examiner Reports — January 2023, June 2023, January 2024, June 2024. Last reviewed 2026-05-12.

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Step-by-step worked examples — Projectiles

Step-by-step solutions to past-paper-style questions on projectiles, written exactly the way a tutor would explain them at the board.

1Range and time of flight on level ground (7 marks, M2)
Core• Adapted from WME02/01 January 2024 Q3• projectile, level ground, range, time of flight
Question
A particle is projected from a point $O$ on horizontal ground with speed $28\,\text{m s}^{-1}$ at an angle of $30^{\circ}$ above the horizontal. Taking $g = 9.8\,\text{m s}^{-2}$ , find (a) the time of flight, (b) the horizontal range.
Step-by-step solution
1. Step 1
  Draw a force/velocity diagram. Resolve the initial velocity into components: horizontal $u_x = 28\cos 30^{\circ}$ , vertical $u_y = 28\sin 30^{\circ} = 14$ .
  $u_x = 28\cos 30^{\circ} = 14\sqrt{3} \approx 24.25\,\text{m s}^{-1}, \quad u_y = 14\,\text{m s}^{-1}$
2. Step 2
  Vertical motion (take upward positive): $a = -9.8$ , $u = 14$ , $s = 0$ at landing. Use $s = ut + \tfrac{1}{2}at^{2}$ .
  $0 = 14t - \tfrac{1}{2}(9.8)t^{2} = t(14 - 4.9t)$
3. Step 3
  Reject $t = 0$ (the start). Solve $14 - 4.9t = 0$ for the time of flight.
  $t = \dfrac{14}{4.9} = \dfrac{20}{7} \approx 2.857\,\text{s}$
4. Step 4
  Horizontal range: $R = u_x \cdot t = 14\sqrt{3} \cdot \dfrac{20}{7} = 40\sqrt{3}\,\text{m} \approx 69.3\,\text{m}$ .
Answer
(a) Time of flight $T = \dfrac{20}{7} \approx 2.86\,\text{s}$ . (b) Range $R = 40\sqrt{3} \approx 69.3\,\text{m}$ .
Examiner tip
M1 for resolving the initial velocity. A1 for both components. M1 for applying $s = ut + \tfrac{1}{2}at^{2}$ vertically. A1 for the correct time. M1 for $R = u_x T$ . A1 for the numerical range. B1 for $3$ s.f. or exact form. Common error: students take downward as positive but forget to flip the sign on $u_y$ . Diagram saves 2 marks reliably.
2Greatest height and trajectory equation (8 marks, M2)
Extended• Adapted from WME02/01 June 2024 Q4• projectile, greatest height, trajectory equation
Question
A stone is projected from ground level with speed $20\,\text{m s}^{-1}$ at an angle $\theta$ above the horizontal, where $\tan\theta = \dfrac{3}{4}$ . Taking $g = 9.8\,\text{m s}^{-2}$ , find (a) the greatest height reached, (b) the equation of the trajectory in the form $y = ax - bx^{2}$ .
Step-by-step solution
1. Step 1
  From $\tan\theta = 3/4$ (a $3$ - $4$ - $5$ triangle), $\sin\theta = 3/5$ and $\cos\theta = 4/5$ . Hence $u_x = 20 \cdot 4/5 = 16$ , $u_y = 20 \cdot 3/5 = 12$ .
2. Step 2
  At greatest height, vertical velocity $= 0$ . Use $v^{2} = u^{2} + 2as$ vertically with $v = 0$ , $u = 12$ , $a = -9.8$ .
  $0 = 12^{2} - 2(9.8)s_{\max} \Rightarrow s_{\max} = \dfrac{144}{19.6} = \dfrac{360}{49} \approx 7.35\,\text{m}$
3. Step 3
  For the trajectory: at time $t$ , $x = 16t$ so $t = x/16$ . Vertically $y = 12t - \tfrac{1}{2}(9.8)t^{2} = 12t - 4.9t^{2}$ .
4. Step 4
  Substitute $t = x/16$ :
  $y = 12 \cdot \dfrac{x}{16} - 4.9 \cdot \dfrac{x^{2}}{256} = \dfrac{3x}{4} - \dfrac{4.9 x^{2}}{256}$
5. Step 5
  Simplify the second coefficient: $4.9 / 256 = 49/2560$ . So $y = \dfrac{3x}{4} - \dfrac{49 x^{2}}{2560}$ , i.e. $a = 3/4$ and $b = 49/2560$ .
Answer
(a) Greatest height $= \dfrac{360}{49} \approx 7.35\,\text{m}$ . (b) $y = \dfrac{3x}{4} - \dfrac{49 x^{2}}{2560}$ .
Examiner tip
M1 for components from $\tan\theta = 3/4$ . A1 for $u_x = 16$ , $u_y = 12$ . M1 for $v^{2} = u^{2} + 2as$ at peak. A1 for $h \approx 7.35$ m. M1 for eliminating $t$ . M1 for substituting into $y$ . A1 for the trajectory equation. The two M marks for the trajectory require a clean substitution — students who try to derive $y = x\tan\theta - \dfrac{gx^{2}}{2u^{2}\cos^{2}\theta}$ from memory often slip on the cosine-squared.
3Projection from a cliff onto sea below (9 marks, M2)
Extended• Adapted from WME02/01 January 2023 Q5• projectile, projection from height
Question
A stone is thrown from the top of a cliff $40\,\text{m}$ above the sea with speed $15\,\text{m s}^{-1}$ at an angle of $\alpha$ below the horizontal, where $\sin\alpha = 3/5$ . Taking $g = 9.8\,\text{m s}^{-2}$ , find (a) the time taken to hit the sea, (b) the horizontal distance travelled, (c) the speed at impact.
Step-by-step solution
1. Step 1
  Set up coordinates with origin at the point of projection, $x$ horizontal in the direction of motion, $y$ measured upward (so the sea is at $y = -40$ ). $\sin\alpha = 3/5$ gives $\cos\alpha = 4/5$ . Initial velocity components: $u_x = 15 \cdot 4/5 = 12$ , $u_y = -15 \cdot 3/5 = -9$ (downward).
2. Step 2
  Vertical motion: $s = -40$ , $u = -9$ , $a = -9.8$ . Use $s = ut + \tfrac{1}{2}at^{2}$ :
  $-40 = -9t - 4.9 t^{2} \Rightarrow 4.9 t^{2} + 9t - 40 = 0$
3. Step 3
  Quadratic formula: $t = \dfrac{-9 + \sqrt{81 + 4 \cdot 4.9 \cdot 40}}{9.8} = \dfrac{-9 + \sqrt{865}}{9.8}$ .
  $t = \dfrac{-9 + 29.41}{9.8} \approx 2.083\,\text{s}$
4. Step 4
  Horizontal distance: $x = u_x t = 12 \cdot 2.083 \approx 25.0\,\text{m}$ .
5. Step 5
  Vertical velocity at impact: $v_y = u_y + at = -9 - 9.8 \cdot 2.083 \approx -29.41\,\text{m s}^{-1}$ . Horizontal velocity is unchanged: $v_x = 12$ .
6. Step 6
  Speed at impact:
  $|v| = \sqrt{12^{2} + 29.41^{2}} = \sqrt{144 + 865} = \sqrt{1009} \approx 31.8\,\text{m s}^{-1}$
Answer
(a) $t \approx 2.08\,\text{s}$ . (b) $\approx 25.0\,\text{m}$ . (c) Impact speed $\approx 31.8\,\text{m s}^{-1}$ .
Examiner tip
M1 for components (note the negative on $u_y$ ). A1 for both. M1 for $s = ut + \tfrac{1}{2}at^{2}$ vertically. A1 for the quadratic. M1 for solving (quadratic formula or calculator). A1 for time. B1 for horizontal distance. M1 for combining $v_x$ and $v_y$ at impact. A1 for impact speed. ECF runs through cleanly from a wrong $t$ . Energy method (KE-gain $=$ PE-loss) also works for (c): $\tfrac{1}{2}v^{2} = \tfrac{1}{2}(15)^{2} + 9.8 \cdot 40 \Rightarrow v \approx 31.8$ — strongly recommended cross-check.
4Projectile passing through a given point — find angle (8 marks, M2)
Extended• Adapted from WME02/01 June 2023 Q5• projectile, trajectory, find angle
Question
A particle is projected from the origin with speed $30\,\text{m s}^{-1}$ at an angle $\theta$ above the horizontal. It passes through the point $(20, 5)$ measured in metres. Taking $g = 9.8\,\text{m s}^{-2}$ , find the two possible values of $\theta$ .
Step-by-step solution
1. Step 1
  Use the trajectory equation $y = x\tan\theta - \dfrac{gx^{2}}{2u^{2}\cos^{2}\theta}$ . Apply $\sec^{2}\theta = 1 + \tan^{2}\theta$ to put everything in terms of $T = \tan\theta$ .
2. Step 2
  Substitute $u = 30$ , $x = 20$ , $y = 5$ , $g = 9.8$ :
  $5 = 20T - \dfrac{9.8 \cdot 400}{2 \cdot 900}(1 + T^{2}) = 20T - \dfrac{49}{22.5}(1 + T^{2})$
3. Step 3
  Simplify: $\dfrac{9.8 \cdot 400}{1800} = \dfrac{3920}{1800} = \dfrac{98}{45} \approx 2.178$ . So:
  $5 = 20T - \dfrac{98}{45}(1 + T^{2})$
4. Step 4
  Multiply by $45$ : $225 = 900T - 98(1 + T^{2})$ , hence $98 T^{2} - 900 T + 323 = 0$ .
5. Step 5
  Quadratic formula:
  $T = \dfrac{900 \pm \sqrt{900^{2} - 4 \cdot 98 \cdot 323}}{2 \cdot 98} = \dfrac{900 \pm \sqrt{810000 - 126616}}{196} = \dfrac{900 \pm \sqrt{683384}}{196}$
6. Step 6
  $\sqrt{683384} \approx 826.67$ . So $T \approx \dfrac{900 + 826.67}{196} \approx 8.81$ or $T \approx \dfrac{900 - 826.67}{196} \approx 0.374$ . Hence $\theta \approx \arctan(8.81) \approx 83.5^{\circ}$ or $\theta \approx \arctan(0.374) \approx 20.5^{\circ}$ .
Answer
$\theta \approx 20.5^{\circ}$ (low trajectory) or $\theta \approx 83.5^{\circ}$ (high trajectory).
Examiner tip
M1 for quoting/deriving the trajectory equation. M1 for substituting $\sec^{2}\theta = 1 + \tan^{2}\theta$ . A1 for the quadratic in $T = \tan\theta$ . M1 for solving the quadratic. A2 for both angles (lose one for missing the second root). The physical interpretation — two angles always reach the same point unless it lies on the maximum-range parabola — is worth highlighting.
5Projection up an inclined plane (9 marks, M2)
Challenge• Adapted from WME02/01 January 2024 Q7• projectile, inclined plane
Question
A particle is projected from a point $A$ on a plane inclined at $20^{\circ}$ to the horizontal. The particle is projected with speed $25\,\text{m s}^{-1}$ at an angle of $40^{\circ}$ above the plane (so $60^{\circ}$ above the horizontal). Taking $g = 9.8\,\text{m s}^{-2}$ , find the distance along the plane from $A$ to the point of landing.
Step-by-step solution
1. Step 1
  Use axes along the plane (call this $x$ ) and perpendicular to the plane ( $y$ ). The acceleration $g$ resolved into these axes is: $g_x = -g\sin 20^{\circ}$ (down the plane), $g_y = -g\cos 20^{\circ}$ (into the plane).
2. Step 2
  Initial velocity along the plane: $u_x = 25\cos 40^{\circ} \approx 19.15$ . Perpendicular: $u_y = 25\sin 40^{\circ} \approx 16.07$ .
3. Step 3
  Particle lands when perpendicular displacement $= 0$ . Use $s = ut + \tfrac{1}{2}at^{2}$ in the $y$ -direction:
  $0 = 16.07 t - \tfrac{1}{2}(9.8\cos 20^{\circ}) t^{2}$
4. Step 4
  $9.8\cos 20^{\circ} \approx 9.209$ , so $0 = 16.07 t - 4.604 t^{2} = t(16.07 - 4.604 t)$ . Reject $t = 0$ : $t = 16.07 / 4.604 \approx 3.491\,\text{s}$ .
5. Step 5
  Distance along the plane: $s_x = u_x t + \tfrac{1}{2}(-g\sin 20^{\circ}) t^{2}$ .
  $s_x = 19.15 \cdot 3.491 - \tfrac{1}{2}(9.8 \sin 20^{\circ})(3.491)^{2}$
6. Step 6
  $9.8\sin 20^{\circ} \approx 3.352$ , $(3.491)^{2} \approx 12.19$ .
  $s_x \approx 66.85 - \tfrac{1}{2} \cdot 3.352 \cdot 12.19 \approx 66.85 - 20.43 \approx 46.4\,\text{m}$
Answer
Distance up the plane $\approx 46.4\,\text{m}$ .
Examiner tip
B1 for choosing axes along/perpendicular to the plane (a diagram is essential and earns this mark). M1 for resolving gravity along and perpendicular to the plane. M1 for resolving velocity along and perpendicular. M1 for time-of-flight from perpendicular displacement $= 0$ . A1 for time. M1 for distance along plane. A1 for the numerical answer. Alternative: standard $(x,y)$ axes also work but the algebra is uglier — the rotated axes are the marker-preferred route.

Key Formulae — Projectiles

The formulae you need to memorise for projectiles on the Pearson Edexcel IAL XMA01 / YMA01 paper, with every variable defined in plain English and a note on when to use it.

Velocity components
$u_x = u\cos\theta, \quad u_y = u\sin\theta$
$u$
initial speed (m s $^{-1}$ )
$\theta$
angle of projection above the horizontal
$u_x$
horizontal component of initial velocity
$u_y$
vertical component of initial velocity
When to use
First step of every projectile problem on horizontal ground. NOT in the IAL formula booklet — but the suvat formulae used afterwards are standard.
Example
$u = 20$ , $\theta = 30^{\circ}$ : $u_x = 20\cos 30^{\circ} = 10\sqrt{3}$ , $u_y = 10$ .
Constant-acceleration formulae (suvat)
$v = u + at, \quad s = ut + \tfrac{1}{2}at^{2}, \quad v^{2} = u^{2} + 2as, \quad s = \tfrac{1}{2}(u + v)t$
$s$
displacement (m)
$u$
initial velocity (m s $^{-1}$ )
$v$
final velocity (m s $^{-1}$ )
$a$
acceleration (m s $^{-2}$ ) — vertical: $\pm g$ ; horizontal: $0$
$t$
time (s)
When to use
Applied separately to horizontal and vertical components. Vertical $a = -g$ (upward positive). Horizontal $a = 0$ so $s = u_x t$ . In the IAL formula booklet under 'kinematics'.
Example
Time to greatest height: set $v = 0$ in $v = u\sin\theta - gt$ to get $t = \dfrac{u\sin\theta}{g}$ .
Range on level ground
$R = \dfrac{u^{2}\sin 2\theta}{g}$
$R$
horizontal range (m)
$u$
initial speed (m s $^{-1}$ )
$\theta$
angle of projection above horizontal
$g$
gravitational acceleration ( $9.8$ or $10$ m s $^{-2}$ )
When to use
Quick range calculation for a particle projected from and returning to the same horizontal level. Maximum at $\theta = 45^{\circ}$ . NOT in the IAL formula booklet — derivable but worth memorising.
Example
$u = 28$ , $\theta = 30^{\circ}$ , $g = 9.8$ : $R = \dfrac{784 \sin 60^{\circ}}{9.8} = 80\sin 60^{\circ} = 40\sqrt{3} \approx 69.3$ m.
Greatest height (level ground)
$H = \dfrac{u^{2}\sin^{2}\theta}{2g}$
$H$
greatest height above projection point (m)
$u, \theta, g$
as above
When to use
When asked for the peak height directly. Derived from $v^{2} = u^{2} + 2as$ with $v_y = 0$ at the top. NOT in the formula booklet.
Example
$u = 20$ , $\theta = 60^{\circ}$ , $g = 9.8$ : $H = \dfrac{400 \cdot 0.75}{19.6} \approx 15.3$ m.
Trajectory (Cartesian path)
$y = x\tan\theta - \dfrac{gx^{2}}{2 u^{2} \cos^{2}\theta}$
$x, y$
horizontal and vertical position from launch (m)
$u, \theta, g$
as above
When to use
When the question gives a point the particle passes through and asks for $\theta$ , or asks for the path equation. Apply $\sec^{2}\theta = 1 + \tan^{2}\theta$ to get a quadratic in $\tan\theta$ . NOT in the formula booklet — derivation expected if asked.
Example
$u = 30$ at $\theta$ passing through $(20, 5)$ : $5 = 20\tan\theta - \dfrac{9.8 \cdot 400}{2 \cdot 900}\sec^{2}\theta$ .
Time of flight on level ground
$T = \dfrac{2u\sin\theta}{g}$
$T$
total time in the air (s)
$u, \theta, g$
as above
When to use
When the particle starts and ends at the same vertical level. For projection from a height, use suvat with $s$ = the height (with appropriate sign). NOT in the formula booklet.
Example
$u = 28$ , $\theta = 30^{\circ}$ , $g = 9.8$ : $T = \dfrac{2 \cdot 14}{9.8} = \dfrac{20}{7} \approx 2.86$ s.

Key Definitions and Keywords — Projectiles

Definitions to memorise and the exact keywords mark schemes credit for projectiles answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IAL XMA01 / YMA01 sitting.

Projectile
Examiner keyword
A particle moving freely under gravity alone (no air resistance, no other forces). Its trajectory is a parabola in any vertical plane containing the initial velocity.
Time of flight
Examiner keyword
The total time the projectile spends in the air, from launch to landing. On level ground, $T = \dfrac{2u\sin\theta}{g}$ .
Range
Examiner keyword
The horizontal distance from the point of projection to the point of landing. On level ground, $R = \dfrac{u^{2}\sin 2\theta}{g}$ , maximised at $\theta = 45^{\circ}$ .
Greatest height
The maximum vertical displacement above the launch point. Reached when $v_y = 0$ , at time $t = \dfrac{u\sin\theta}{g}$ .
Trajectory (path equation)
The Cartesian equation $y$ as a function of $x$ describing the projectile's path. Obtained by eliminating $t$ from $x = u\cos\theta \cdot t$ and $y = u\sin\theta \cdot t - \tfrac{1}{2}gt^{2}$ .
Angle of projection
Examiner keyword
The angle $\theta$ above the horizontal at which the projectile is launched. Two distinct angles ( $\theta$ and $90^{\circ} - \theta$ ) give the same range on level ground.

Common Mistakes and Misconceptions — Projectiles

The traps other students keep falling into on projectiles questions — taken from recent Pearson Edexcel IAL XMA01 / YMA01 examiner reports and mark schemes — and how to avoid them.

✕Using $a = +g$ for vertical motion when upward is taken positive
WME02/01 examiner reports — flagged annually
Why it happens
Students see $9.8$ as 'the acceleration' and forget gravity acts downward.
How to avoid it
State your sign convention at the top of the working: 'taking upward as positive, $a = -9.8\,\text{m s}^{-2}$ '. Mark schemes credit clear conventions and apply ECF if the convention is consistent.
✕Treating the speed $u$ as both horizontal and vertical components
Why it happens
Forgetting that velocity is a vector and must be decomposed.
How to avoid it
First line of every projectile solution: $u_x = u\cos\theta$ , $u_y = u\sin\theta$ . Draw the right-angled triangle of components.
✕Writing the trajectory equation as $y = x\tan\theta - \dfrac{gx^{2}}{2u^{2}}$ (missing the $\cos^{2}\theta$ )
WME02/01 examiner reports — recurring
Why it happens
Memorisation slip — students confuse it with the simpler vertical-motion formula.
How to avoid it
Re-derive: from $x = u\cos\theta \cdot t$ , $t = x/(u\cos\theta)$ . Substitute into $y = u\sin\theta \cdot t - \tfrac{1}{2}g t^{2}$ . The $\cos^{2}\theta$ comes from $t^{2}$ .
✕Using horizontal/vertical axes for projection on an inclined plane
Why it happens
Default habit from level-ground problems.
How to avoid it
On an inclined plane, ALWAYS use axes along and perpendicular to the plane. The 'landing condition' is then simply perpendicular displacement $= 0$ , exactly as on level ground.
✕Taking the vertical-velocity-at-impact as the impact speed
Why it happens
Forgetting the horizontal component is still there.
How to avoid it
Impact speed $= \sqrt{v_x^{2} + v_y^{2}}$ . $v_x$ is constant throughout the flight (no air resistance); $v_y$ changes. Always combine.

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

Get a report showing which sub-topics you've nailed and which ones still need work.

Video lesson

Short walkthrough of the concepts students most often get stuck on.

Projectiles — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Projectiles

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1Range and time of flight on level ground (7 marks, M2)

2Greatest height and trajectory equation (8 marks, M2)

3Projection from a cliff onto sea below (9 marks, M2)

4Projectile passing through a given point — find angle (8 marks, M2)

5Projection up an inclined plane (9 marks, M2)

Velocity components

Constant-acceleration formulae (suvat)

Range on level ground

Greatest height (level ground)

Trajectory (Cartesian path)

Time of flight on level ground

Projectile

Time of flight

Range

Greatest height

Trajectory (path equation)

Angle of projection

✕Using a=+ga = +ga=+g for vertical motion when upward is taken positive

✕Treating the speed uuu as both horizontal and vertical components

✕Writing the trajectory equation as y=xtan⁡θ−gx22u2y = x\tan\theta - \dfrac{gx^{2}}{2u^{2}}y=xtanθ−2u2gx2​ (missing the cos⁡2θ\cos^{2}\thetacos2θ)

✕Using horizontal/vertical axes for projection on an inclined plane

✕Taking the vertical-velocity-at-impact as the impact speed

Practice questions

Past paper style quiz

Video lesson

Projectiles — frequently asked questions

✕Using $a = +g$ for vertical motion when upward is taken positive

✕Treating the speed $u$ as both horizontal and vertical components

✕Writing the trajectory equation as $y = x\tan\theta - \dfrac{gx^{2}}{2u^{2}}$ (missing the $\cos^{2}\theta$ )