Projectile Motion

A projectile is any object that has been launched and then moves only under the influence of gravity. A football kicked into the air, a bullet fired from a gun, a stone thrown off a cliff — all are projectiles once they leave the launcher. Understanding projectile motion is a key skill for Edexcel A-Level Physics and combines vector resolution with SUVAT equations.

The Key Principle: Independence of Horizontal and Vertical Motion

The most important idea in projectile motion is that horizontal and vertical motions are independent of each other. This means:

The horizontal velocity has no effect on the vertical velocity, and vice versa.
Gravity acts only in the vertical direction, so it affects only the vertical component of motion.
The horizontal component of velocity remains constant (assuming no air resistance).
The vertical component of velocity changes at a rate of g = 9.81 m/s² downward.

This allows us to treat the two directions completely separately, solving each with SUVAT equations independently, then combining the results.

flowchart TD
    A["Projectile launched at speed u at angle θ"] --> B["Resolve initial velocity"]
    B --> C["Horizontal: uₓ = u cos θ"]
    B --> D["Vertical: uᵧ = u sin θ"]
    C --> E["Horizontal motion\naₓ = 0 (constant velocity)\nx = uₓ t"]
    D --> F["Vertical motion\naᵧ = -g (SUVAT applies)\ny = uᵧ t - ½gt²"]
    E --> G["Combine results\nusing shared time t"]
    F --> G

Resolving the Initial Velocity

If a projectile is launched at speed u at angle theta to the horizontal:

Horizontal component: u_x = u cos(theta)
Vertical component: u_y = u sin(theta)

Once resolved, the horizontal motion has constant velocity u_x, and the vertical motion has initial velocity u_y with acceleration g downward.

Quick Reference: Component Values for Common Angles

Launch Angle	cos(theta)	sin(theta)	u_x (as fraction of u)	u_y (as fraction of u)
0° (horizontal)	1.000	0.000	u	0
30°	0.866	0.500	0.866u	0.500u
45°	0.707	0.707	0.707u	0.707u
60°	0.500	0.866	0.500u	0.866u
90° (vertical)	0.000	1.000	0	u

Horizontal Motion (constant velocity)

Since there is no horizontal acceleration (ignoring air resistance):

Horizontal displacement: x = u_x t = (u cos theta) t

This is simply distance = speed x time.

Vertical Motion (uniform acceleration under gravity)

Using SUVAT with a = -g (taking upward as positive):

v_y = u_y - gt
y = u_y t - (1/2)gt²
v_y² = u_y² - 2gy

Time of Flight

The time of flight is the total time the projectile is in the air. For a projectile launched from and landing at the same height, the vertical displacement is zero when it lands.

Using y = u_y t - (1/2)gt²: 0 = u_y t - (1/2)gt² 0 = t(u_y - (1/2)gt)

This gives t = 0 (launch) and t = 2u_y / g = 2u sin(theta) / g (landing).

Time of flight = 2u sin(theta) / g

Maximum Height

At the highest point, the vertical velocity is zero (v_y = 0).

Using v_y² = u_y² - 2gy: 0 = u_y² - 2g H H = u_y² / (2g) = u² sin²(theta) / (2g)

Maximum height = u² sin²(theta) / (2g)

Range

The range is the horizontal distance travelled during the time of flight.

Range = u_x x time of flight = u cos(theta) x 2u sin(theta) / g = u² sin(2 theta) / g

Range = u² sin(2 theta) / g

This formula shows that the maximum range occurs when sin(2 theta) = 1, i.e. theta = 45° (for level ground with no air resistance).

Important: Complementary angles (e.g. 30° and 60°) give the same range because sin(2 x 30°) = sin(60°) and sin(2 x 60°) = sin(120°) = sin(60°).

Summary of Key Projectile Formulae (Level Ground)

Quantity	Formula	Notes
Time of flight	T = 2u sin(theta) / g	Only for same launch/landing height
Maximum height	H = u² sin²(theta) / (2g)	Occurs at t = T/2
Range	R = u² sin(2 theta) / g	Maximum at theta = 45°
Speed at top	v = u cos(theta)	Horizontal component only

Worked Example 1: Projectile Launched at an Angle

A ball is kicked at 20 m/s at 30° above the horizontal from ground level. Find the time of flight, maximum height, and range. (g = 9.81 m/s²)

Resolve initial velocity: u_x = 20 cos(30°) = 20 x 0.866 = 17.3 m/s u_y = 20 sin(30°) = 20 x 0.500 = 10.0 m/s

Time of flight: T = 2u_y / g = 2 x 10.0 / 9.81 = 20.0 / 9.81 = 2.04 s

Maximum height: H = u_y² / (2g) = 10.0² / (2 x 9.81) = 100 / 19.62 = 5.10 m

Range: R = u_x x T = 17.3 x 2.04 = 35.3 m

Worked Example 2: Horizontal Launch from a Cliff

A stone is thrown horizontally at 15 m/s from a cliff 80 m high. How long does it take to reach the ground, and how far from the base of the cliff does it land?

The initial vertical velocity is zero (horizontal launch): u_y = 0.

Vertical motion (taking downward as positive): s = u_y t + (1/2)gt² 80 = 0 + (1/2)(9.81)t² t² = 160 / 9.81 = 16.31 t = 4.04 s

Horizontal distance: x = u_x t = 15 x 4.04 = 60.6 m

The stone lands 60.6 m from the base of the cliff.

Worked Example 3: Finding the Speed at Impact

Using the same cliff problem, find the speed of the stone just before it hits the ground.

Horizontal velocity at impact: v_x = u_x = 15 m/s (constant)

Vertical velocity at impact: v_y = u_y + gt = 0 + 9.81 x 4.04 = 39.6 m/s

Speed = sqrt(v_x² + v_y²) = sqrt(15² + 39.6²) = sqrt(225 + 1568) = sqrt(1793) = 42.3 m/s

Angle below horizontal: theta = arctan(39.6 / 15) = arctan(2.64) = 69.3°

Worked Example 4: Projectile Launched Below the Horizontal

A ball is thrown at 12 m/s at 20° below the horizontal from the top of a 30 m building. Find the time to reach the ground and the horizontal distance.

u_x = 12 cos(20°) = 12 x 0.940 = 11.3 m/s u_y = -12 sin(20°) = -12 x 0.342 = -4.10 m/s (negative because downward)

Taking downward as positive and the top of the building as origin: 30 = 4.10t + (1/2)(9.81)t² 4.905t² + 4.10t - 30 = 0

Using the quadratic formula: t = (-4.10 + sqrt(4.10² + 4 x 4.905 x 30)) / (2 x 4.905) t = (-4.10 + sqrt(16.81 + 588.6)) / 9.81 t = (-4.10 + sqrt(605.4)) / 9.81 t = (-4.10 + 24.6) / 9.81 = 20.5 / 9.81 = 2.09 s

Horizontal distance: x = 11.3 x 2.09 = 23.6 m

Effects of Air Resistance

In examination problems, air resistance is usually neglected. However, it is important to understand qualitatively what happens when air resistance is present:

Air resistance opposes the direction of motion, so it reduces both horizontal and vertical speeds.
The horizontal velocity is no longer constant — it decreases over time.
The trajectory becomes asymmetric: the descent is steeper than the ascent.
The maximum height and range are both reduced.
The time of flight may increase slightly because the descent is slower.

The shape of the actual trajectory shifts from a symmetric parabola to an asymmetric curve that falls more steeply.

Common Mistakes and Misconceptions

Forgetting to resolve the initial velocity into components before applying SUVAT.
Mixing up horizontal and vertical: acceleration only acts vertically; horizontal velocity is constant.
Sign errors: if upward is positive, gravity is negative. Be consistent.
Using range/time of flight formulas when the launch and landing heights differ — these formulas assume the same height. For different heights, solve the vertical SUVAT equation directly.
Thinking the velocity is zero at the top: The vertical component is zero, but the horizontal component is unchanged. The speed at the top equals u cos(theta), not zero.
Forgetting the quadratic formula: When launch and landing heights differ, the vertical equation is a quadratic in t. You must use the quadratic formula and take the positive root.

Exam Tips for Edexcel Physics

Always start by resolving the initial velocity into horizontal and vertical components and write these down clearly.
State your sign convention (e.g. "taking upward as positive").
For cliff problems where the projectile lands below the launch point, make sure the vertical displacement has the correct sign.
If the question asks for speed (not velocity), remember to combine v_x and v_y using Pythagoras.
If asked for the angle of impact, use arctan(v_y / v_x).
Draw a quick sketch of the trajectory — this helps identify which components to use.

Key Points

Horizontal and vertical motions are independent and can be analysed separately.
Horizontal: constant velocity (no acceleration). Vertical: uniform acceleration under gravity.
Resolve the initial velocity into horizontal and vertical components using cosine and sine.
Time of flight, maximum height, and range can all be found using SUVAT equations applied to the appropriate direction.
Complementary launch angles give the same range on level ground.
Air resistance reduces range and maximum height, and makes the trajectory asymmetric.
When launch and landing heights differ, solve the quadratic equation for time directly.

A-Level Deep Dive: Projectile Motion

Spec mapping

Edexcel 9PH0 specification, Topic 2 — Mechanics, sub-topic 2.2 (Motion) requires candidates to analyse the independent horizontal and vertical components of motion of a projectile, applying the equations of uniformly accelerated motion in each direction (refer to the official Pearson Edexcel specification document for exact wording). Although projectile motion is first formalised here, the same component-decomposition technique recurs across sub-topic 2.1 (Scalars and vectors) for the underlying resolution of the launch velocity, sub-topic 2.3 (Forces) when modelling oblique impact and rebound, Topic 6 (Further mechanics) where two-dimensional momentum conservation governs glancing collisions, Topic 7 (Electric and magnetic fields) for charged-particle deflection in uniform fields (the parabolic path of an electron between deflecting plates is a literal projectile problem with electric weight), and Topic 8 (Nuclear and particle physics) for cloud-chamber track interpretation. The Edexcel data, formulae and relationships booklet supplies the suvat list but does not state $R = (u^2\sin 2\theta)/g$ or $H = (u^2\sin^2\theta)/(2g)$ — these range and maximum-height shortcuts must be derived from suvat in working.

Worked example with full mark scheme

Question (8 marks):

A ball is launched from ground level at an angle of 30° above the horizontal with an initial speed of 20.0 m s⁻¹. Air resistance is negligible and $g = 9.81$ m s⁻².

(a) Calculate the horizontal and vertical components of the initial velocity. (2)

(b) Determine the total time of flight before the ball returns to the launch height. (2)

(d) Calculate the range on level ground and state the speed and direction of the velocity at the highest point. (2)

Solution with mark scheme:

(a) Step 1 — resolve the launch velocity. Take rightward as positive $x$ and upward as positive $y$ .

$u_x = u\cos\theta = 20.0\cos 30° = 20.0 \times \dfrac{\sqrt{3}}{2} = 17.3 \text{ m s}^{-1}$

$u_y = u\sin\theta = 20.0\sin 30° = 20.0 \times \dfrac{1}{2} = 10.0 \text{ m s}^{-1}$

M1 — correct resolution using cosine for the horizontal (adjacent) and sine for the vertical (opposite) component, with the angle measured from the horizontal.

A1 — both components correct to 3 sf with units.

(b) Step 1 — vertical motion governs time of flight. With launch and landing heights equal, the vertical displacement over the full flight is zero. Apply $s = u_y t + \tfrac{1}{2}a t^2$ with $s_y = 0$ and $a = -g$ :

$0 = 10.0\,t - \tfrac{1}{2}(9.81)t^2 = t\left(10.0 - 4.905\,t\right)$

M1 — setting $s_y = 0$ and recognising the trivial root $t = 0$ corresponds to launch.

The non-trivial root: $t = 10.0 / 4.905 = 2.04$ s.

A1 — $T = 2.04$ s (3 sf).

$0 = (10.0)^2 + 2(-9.81)H \implies H = \dfrac{100}{19.62} = 5.10 \text{ m}$

M1 — using a vertical-only suvat with $v_y = 0$ at the apex.

A1 — $H = 5.10$ m (3 sf).

(d) Step 1 — range from horizontal-only suvat. Horizontal velocity is constant, so:

$R = u_x \times T = 17.3 \times 2.04 = 35.3 \text{ m}$

M1 — using $R = u_x T$ with the time of flight from (b).

At the apex the vertical component is zero and the horizontal component is unchanged, so the velocity is $17.3$ m s⁻¹ horizontally.

A1 — $R = 35.3$ m (3 sf) and apex velocity stated as 17.3 m s⁻¹ horizontal.

Total: 8 marks (M4 A4).

Specimen question modelled on the Edexcel 9PH0 Paper 1 format

Question (6 marks): A stone is launched horizontally from the top of a cliff 45.0 m above level ground with an initial speed of 12.0 m s⁻¹. Air resistance is negligible and $g = 9.81$ m s⁻².

(a) Calculate the time taken for the stone to reach the ground. (2)

(b) Calculate the horizontal range from the foot of the cliff to the point where the stone lands. (1)

(c) Determine the magnitude and direction (relative to the horizontal) of the velocity with which the stone strikes the ground. (3)

Mark scheme decomposition by AO:

Part	AO weighting	Earned by
(a) M1	AO1.2	Identifying $u_y = 0$ and applying $s_y = \tfrac{1}{2}gt^2$ in the vertical direction.
(a) A1	AO1.1b	$t = \sqrt{2 \times 45.0 / 9.81} = 3.03$ s (3 sf).
(b) A1	AO1.1b	$R = u_x t = 12.0 \times 3.03 = 36.4$ m (3 sf).
(c) M1	AO2.1	Computing the vertical impact speed $v_y = gt = 9.81 \times 3.03 = 29.7$ m s⁻¹.
(c) M1	AO1.2	Combining components by Pythagoras: $v = \sqrt{12.0^2 + 29.7^2} = 32.0$ m s⁻¹.
(c) A1	AO2.1	Direction: $\theta = \tan^{-1}(29.7/12.0) = 68.0°$ below the horizontal, with explicit reference to the chosen sign convention.

Total: 6 marks split AO1 = 4, AO2 = 2. This is a textbook horizontal-launch projectile problem with an AO2 reasoning tail — typical of Paper 1 projectile items, which reward candidates who treat the vertical and horizontal motions as independent and only recombine at the impact instant via vector addition.

Synoptic links

Sub-topic 2.1 (Scalars and vectors): the launch velocity must be resolved into perpendicular components before any suvat reasoning begins. Projectile problems are the canonical synoptic test of vector resolution — every projectile question is a vector question dressed in kinematic clothing.
Sub-topic 2.1 (Kinematics, uniform acceleration): suvat is applied independently in each direction. The horizontal direction has $a = 0$ (so $s_x = u_x t$ ), the vertical has $a = -g$ (full suvat). Pattern recognition: the same five variables per axis; the trick is choosing axes so one component is trivially constant.
Sub-topic 4.3 (Energy: KE $\leftrightarrow$ PE conversion): at the apex of a projectile, kinetic energy equals $\tfrac{1}{2}mu_x^2$ (only the horizontal component survives), while the gravitational potential energy gained is $\tfrac{1}{2}m u_y^2$ . Energy conservation between launch and apex gives the maximum-height formula $H = u_y^2/(2g)$ without any kinematics — a powerful cross-check.