Projectile Motion

This lesson covers projectile motion as required by the Edexcel 9MA0 A-Level Mathematics specification. A projectile moves in two dimensions under gravity alone — this combines horizontal constant velocity with vertical constant acceleration.

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Assumptions

The standard projectile model assumes:

• The projectile is a particle (no air resistance, no spin).
• Acceleration due to gravity, g, acts vertically downwards and is constant (9.8 m s⁻²).
• The horizontal acceleration is zero — there is no air resistance.

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Resolving the Initial Velocity

If a projectile is launched with speed u at angle θ above the horizontal:

• Horizontal component: ux = u cos θ
• Vertical component: uy = u sin θ

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Equations of Motion

Horizontal (no acceleration)

x = (u cos θ)t

Vertical (constant acceleration g downwards)

Taking upwards as positive:

vy = u sin θ - gt y = (u sin θ)t - ½gt² vy² = (u sin θ)² - 2gy

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Key Results

Time of Flight

The total time for the projectile to return to its launch height (y = 0):

0 = (u sin θ)t - ½gt² t(u sin θ - ½gt) = 0 t = 0 (launch) or t = 2u sin θ / g (landing)

Time of flight T = 2u sin θ / g

Maximum Height

At the highest point, vy = 0: 0 = (u sin θ)² - 2gH H = u² sin²θ / (2g)

This occurs at time t = u sin θ / g (half the time of flight).

Range

The horizontal distance at the time of landing:

R = (u cos θ) x T = u cos θ x 2u sin θ / g = u² sin 2θ / g

The maximum range occurs when sin 2θ = 1, i.e. θ = 45°.

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The Equation of the Trajectory

Eliminating t from the parametric equations:

From x = (u cos θ)t: t = x / (u cos θ)

Substituting into y = (u sin θ)t - ½gt²:

y = x tan θ - gx² / (2u² cos²θ)

This is a quadratic in x, confirming the path is a parabola.

Exam Tip: You do not need to memorise the trajectory equation. If needed, derive it by eliminating t from the horizontal and vertical equations.

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Projectiles Launched from a Height

If the projectile is launched from height h above the ground, the vertical equation becomes:

y = h + (u sin θ)t - ½gt²

The projectile hits the ground when y = 0: 0 = h + (u sin θ)t - ½gt²

Solve this quadratic for t (take the positive root).

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Worked Example

A ball is kicked from ground level at 20 m s⁻¹ at 30° above the horizontal. Find the time of flight, maximum height and range. Take g = 9.8 m s⁻².

ux = 20 cos 30° = 20 x (√3/2) = 10√3 ≈ 17.32 m s⁻¹ uy = 20 sin 30° = 20 x 0.5 = 10 m s⁻¹

Time of flight: T = 2(10)/9.8 = 20/9.8 ≈ 2.04 s

Maximum height: H = 10²/(2 x 9.8) = 100/19.6 ≈ 5.10 m

Range: R = 10√3 x 2.04 ≈ 17.32 x 2.04 ≈ 35.3 m

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Velocity at Any Time

The velocity vector at time t is:

v = (u cos θ)i + (u sin θ - gt)j

The speed at time t is:

|v| = √((u cos θ)² + (u sin θ - gt)²)

The angle the velocity makes with the horizontal:

α = arctan((u sin θ - gt) / (u cos θ))

At the highest point, the velocity is purely horizontal: v = (u cos θ)i, so the speed is u cos θ.

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Summary

• A projectile has zero horizontal acceleration and constant vertical acceleration g downwards.
• Resolve initial velocity into horizontal (u cos θ) and vertical (u sin θ) components.
• Time of flight T = 2u sin θ / g (for ground-to-ground).
• Maximum height H = u² sin²θ / (2g).
• Range R = u² sin 2θ / g, maximised when θ = 45°.
• The trajectory is a parabola.
• For projectiles launched from height h, set y = 0 and solve the quadratic.

A-Level Deep Dive: Projectile Motion

Spec mapping

Edexcel 9MA0-03 specification section 9 — Kinematics (Mechanics) covers model motion under gravity in a vertical plane using vectors; projectiles (refer to the official specification document for exact wording). Projectiles sit in Paper 3 — Statistics and Mechanics, Year 2 content, and assume confident handling of all SUVAT relations from Year 1 kinematics. The model is highly idealised: a particle moves under uniform gravitational acceleration $g = 9.8\,\text{m s}^{-2}$g=9.8m s−2 (the formula booklet quotes $9.8$9.8, not $9.81$9.81), air resistance is neglected, the projectile is treated as a point mass, and the ground is treated as horizontal. Projectile questions intersect with section 8 (Vectors) for resolving initial velocity, section 5 of Paper 1 (Trigonometry) for the angle algebra, and section 7 of Paper 2 (Parametric equations) when the trajectory is described as $x(t)$x(t), $y(t)$y(t).

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Worked example with full mark scheme

Question (8 marks):

A ball is projected from a point $O$O at ground level with speed $U = 28\,\text{m s}^{-1}$U=28m s−1 at an angle $\alpha$α above the horizontal, where $\tan\alpha = \tfrac{3}{4}$tanα=43​. The ball moves freely under gravity and lands on horizontal ground. Take $g = 9.8\,\text{m s}^{-2}$g=9.8m s−2.

(a) Find the time of flight $T$T. (2)

(b) Find the horizontal range $R$R. (2)

(c) Find the maximum height $H$H above $O$O. (2)

(d) Show that the equation of the trajectory is $y = \tfrac{3}{4}x - \dfrac{g x^2}{2 u_x^2}$y=43​x−2ux2​gx2​. (2)

Solution with mark scheme:

Since $\tan\alpha = \tfrac{3}{4}$tanα=43​ in a 3-4-5 triangle, $\sin\alpha = \tfrac{3}{5}$sinα=53​ and $\cos\alpha = \tfrac{4}{5}$cosα=54​. Resolve the initial velocity:

$u_x = U\cos\alpha = 28 \cdot \tfrac{4}{5} = 22.4\,\text{m s}^{-1}, \qquad u_y = U\sin\alpha = 28 \cdot \tfrac{3}{5} = 16.8\,\text{m s}^{-1}$ux​=Ucosα=28⋅54​=22.4m s−1,uy​=Usinα=28⋅53​=16.8m s−1

(a) Step 1 — vertical motion to find $T$T. With $u_y = 16.8$uy​=16.8, $a = -9.8$a=−9.8, $s = 0$s=0 (lands at launch height):

$0 = 16.8\,T - \tfrac{1}{2}(9.8)T^2 \implies T(16.8 - 4.9\,T) = 0$0=16.8T−21​(9.8)T2⟹T(16.8−4.9T)=0

M1 — applying $s = u_y t + \tfrac{1}{2}a t^2$s=uy​t+21​at2 vertically with $s = 0$s=0 and $a = -g$a=−g. Reject $T = 0$T=0 (launch instant).

A1 — $T = \dfrac{16.8}{4.9} = \dfrac{24}{7} \approx 3.43\,\text{s}$T=4.916.8​=724​≈3.43s.

(b) Step 2 — horizontal range. Horizontal motion is constant velocity:

$R = u_x \cdot T = 22.4 \cdot \tfrac{24}{7} = 76.8\,\text{m}$R=ux​⋅T=22.4⋅724​=76.8m

M1 — $R = u_x T$R=ux​T using horizontal constant velocity (no acceleration term).

A1 — $R = 76.8\,\text{m}$R=76.8m.

(c) Step 3 — maximum height. At max height $v_y = 0$vy​=0. Using $v_y^2 = u_y^2 + 2as$vy2​=uy2​+2as vertically:

$0 = 16.8^2 + 2(-9.8)H \implies H = \dfrac{16.8^2}{2 \cdot 9.8} = 14.4\,\text{m}$0=16.82+2(−9.8)H⟹H=2⋅9.816.82​=14.4m

M1 — recognising $v_y = 0$vy​=0 at max height and applying $v^2 = u^2 + 2as$v2=u2+2as vertically.

A1 — $H = 14.4\,\text{m}$H=14.4m.

(d) Step 4 — trajectory equation. Eliminate $t$t between $x = u_x t$x=ux​t and $y = u_y t - \tfrac{1}{2}g t^2$y=uy​t−21​gt2. From the first, $t = \dfrac{x}{u_x}$t=ux​x​. Substitute:

$y = u_y \cdot \dfrac{x}{u_x} - \tfrac{1}{2}g \left(\dfrac{x}{u_x}\right)^2 = x \tan\alpha - \dfrac{g x^2}{2 u_x^2} = \tfrac{3}{4}x - \dfrac{g x^2}{2 u_x^2}$y=uy​⋅ux​x​−21​g(ux​x​)2=xtanα−2ux2​gx2​=43​x−2ux2​gx2​

M1 — eliminating $t$t via $t = x / u_x$t=x/ux​ and substituting.

A1 — final form matching the printed result with $\tan\alpha$tanα as the linear coefficient and $\dfrac{g}{2 u_x^2}$2ux2​g​ as the quadratic coefficient.

Total: 8 marks.

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Specimen question modelled on the Edexcel 9MA0 Paper 3 format

Question (6 marks): A stone is thrown from a cliff at height $h = 25\,\text{m}$h=25m above the sea, with initial speed $U = 14\,\text{m s}^{-1}$U=14m s−1 at angle $\alpha$α above the horizontal where $\sin\alpha = \tfrac{3}{7}$sinα=73​. The stone moves freely under gravity ($g = 9.8\,\text{m s}^{-2}$g=9.8m s−2).

(a) Find the time taken to hit the sea. (4)

(b) Find the horizontal distance from the foot of the cliff at which the stone enters the sea. (2)

Mark scheme decomposition by AO:

(a)

• M1 (AO3.1b) — modelling: setting up the vertical equation with $s = -25$s=−25 (downward displacement, taking up positive), $u_y = 14 \cdot \tfrac{3}{7} = 6$uy​=14⋅73​=6, $a = -9.8$a=−9.8.
• M1 (AO1.1b) — applying $s = ut + \tfrac{1}{2}at^2$s=ut+21​at2: $-25 = 6t - 4.9 t^2$−25=6t−4.9t2.
• M1 (AO1.1a) — rearranging to standard quadratic form $4.9 t^2 - 6t - 25 = 0$4.9t2−6t−25=0.
• A1 (AO1.1b) — solving and selecting the positive root: $t = \dfrac{6 + \sqrt{36 + 490}}{9.8} = \dfrac{6 + \sqrt{526}}{9.8} \approx 2.95\,\text{s}$t=9.86+36+490​​=9.86+526​​≈2.95s.

(b)

• M1 (AO1.1b) — horizontal distance $x = u_x t$x=ux​t with $u_x = 14\cos\alpha = 14\sqrt{1 - 9/49} = 4\sqrt{10}$ux​=14cosα=141−9/49​=410​.
• A1 (AO2.5) — $x \approx 4\sqrt{10} \cdot 2.95 \approx 37.3\,\text{m}$x≈410​⋅2.95≈37.3m.

Total: 6 marks split AO1 = 4, AO2 = 1, AO3 = 1.

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Synoptic links

Connects to:

1. Year 1 kinematics — SUVAT applied independently in each direction: projectile motion is the canonical application of SUVAT, but with the crucial twist that the constant-acceleration equations are applied separately to the horizontal ($a = 0$a=0) and vertical ($a = -g$a=−g) components. Forgetting that horizontal motion has zero acceleration is the most common Year 1 $\to$→ Year 2 error.
2. Section 8 — Vectors: the initial velocity is a vector $\mathbf{u} = U(\cos\alpha\,\mathbf{i} + \sin\alpha\,\mathbf{j})$u=U(cosαi+sinαj) and the position vector at time $t$t is $\mathbf{r}(t) = \mathbf{u}t + \tfrac{1}{2}\mathbf{g}t^2$r(t)=ut+21​gt2 where $\mathbf{g} = -g\,\mathbf{j}$g=−gj. Edexcel sometimes phrases projectile questions entirely in vector form.
3. Paper 1 section 5 — Trigonometry: resolving $U$U into $U\cos\alpha$Ucosα and $U\sin\alpha$Usinα relies on confident SOH-CAH-TOA, and the trajectory derivation uses $\sec^2\alpha = 1 + \tan^2\alpha$sec2α=1+tan2α when written compactly as $y = x\tan\alpha - \dfrac{g x^2 \sec^2\alpha}{2 U^2}$y=xtanα−2U2gx2sec2α​.
4. Paper 2 section 7 — Parametric equations: the trajectory $\{x(t), y(t)\}${x(t),y(t)} is a parametric curve with parameter $t$t. Eliminating $t$t to obtain a Cartesian equation $y = f(x)$y=f(x) is the standard parametric-to-Cartesian conversion.
5. Modelling extension — air resistance: introducing a drag term $-k\mathbf{v}$−kv converts the constant-acceleration ODE into $m\dot{\mathbf{v}} = m\mathbf{g} - k\mathbf{v}$mv˙=mg−kv, a first-order linear system. This is beyond 9MA0 but is the natural undergraduate extension.

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Mark-scheme literacy

Projectile questions on 9MA0-03 split AO marks more evenly than Paper 1 algebra questions:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	50–60%	Resolving velocity, applying SUVAT in each direction, executing the algebra
AO2 (reasoning / interpretation)	20–30%	Justifying $v_y = 0$vy​=0 at max height, selecting the positive root, interpreting "lands" as $y = 0$y=0 or $y = -h$y=−h
AO3 (modelling / problem-solving)	15–25%	Setting up the model, choosing axes, deciding which SUVAT equation, treating the cliff geometry

Examiner-rewarded phrasing: "resolving the initial velocity into perpendicular components"; "since horizontal acceleration is zero, $x = u_x t$x=ux​t"; "at maximum height the vertical component of velocity is zero (the horizontal component is unchanged)". Phrases that lose marks: "the velocity is zero at max height" (only the vertical component is); "applying $v = u + at$v=u+at to the whole motion" (which mixes components).

A specific Edexcel pattern: when the projectile lands below its launch point (cliff problem), the displacement $s = -h$s=−h (negative, downward, taking up as positive). Treating it as $+h$+h produces a quadratic with no real roots or wrong sign — and the M-mark for "correct equation" is lost.

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Grade-band model answers

3-mark question

Question: A particle is projected horizontally from a height of $20\,\text{m}$20m with speed $15\,\text{m s}^{-1}$15m s−1. Find the time taken to reach the ground (take $g = 9.8\,\text{m s}^{-2}$g=9.8m s−2).

Grade C response (~190 words):