Projectile Motion: GCSE Physics Guide

A projectile is any object that moves under gravity alone after launch. The horizontal and vertical components of its motion are completely independent: horizontal velocity stays constant while vertical velocity increases at 10 m/s² downward due to gravity. These two facts allow every projectile problem to be solved by treating the two directions separately.

What is a projectile?

A projectile is any object that has been launched into the air and is moving under the influence of gravity only — no engine, no thrust, no air resistance (in GCSE problems). Examples include a kicked football (while in the air), a cannonball, a ball thrown from a window, or a stone dropped from a cliff while walking.

The defining feature of projectile motion is that once the object is in the air:

• No horizontal force acts on it (ignoring air resistance)
• Gravity acts vertically downward throughout the motion

This means the horizontal and vertical motions are controlled by completely different forces — and therefore behave completely independently.

Why are horizontal and vertical components independent?

Newton's first law tells us that an object maintains its velocity unless acted on by a net force. In projectile motion:

• Horizontal: no horizontal force acts → the horizontal velocity is constant throughout the flight. The projectile moves the same horizontal distance every second.
• Vertical: gravity provides a constant downward force → the vertical velocity increases (downward) at a constant rate. At GCSE, gravitational field strength g = 10 m/s² (or 9.8 m/s² in some papers — use whichever is given).

This independence means you can completely analyse the horizontal and vertical motions separately, then combine the results.

How do you calculate the time of flight for a horizontal projectile?

A horizontally launched projectile is one where the initial vertical velocity is zero — the object is launched purely horizontally (thrown off a cliff, rolled off a table, etc.).

For the vertical motion only:

• Initial vertical velocity: u_y = 0 (launched horizontally)
• Acceleration: g = 10 m/s² downward
• Using: s = ½ g t² (where s is the vertical distance fallen)
• Rearranging: t = √(2s ÷ g)

This gives the time of flight, which is entirely determined by the height from which the object was launched.

Worked example: calculating range and landing velocity

Problem: A ball is launched horizontally from the top of a cliff 45 m above the ground. Its horizontal speed is 20 m/s. Calculate: (a) the time of flight (b) the horizontal range (c) the vertical velocity just before it hits the ground

(a) Time of flight:

Use the vertical motion: s = ½ g t²

t = √(2s ÷ g) = √(2 × 45 ÷ 10) = √(90 ÷ 10) = √9 = 3.0 s

(b) Horizontal range:

Horizontal velocity is constant at 20 m/s:

Range = horizontal velocity × time = 20 × 3.0 = 60 m

(c) Vertical velocity at landing:

The initial vertical velocity was 0; gravity accelerates the ball downward at 10 m/s²:

v_y = u_y + g × t = 0 + 10 × 3.0 = 30 m/s downward

The ball hits the ground with a horizontal velocity of 20 m/s and a vertical velocity of 30 m/s. The actual speed at impact can be found using Pythagoras: speed = √(20² + 30²) = √(400 + 900) = √1300 ≈ 36 m/s, at an angle below the horizontal.

What determines how far a projectile travels?

For a horizontal launch, the range depends on:

• The horizontal speed (greater speed → greater range)
• The height of launch (greater height → longer time of flight → greater range)

These two quantities are completely independent. Doubling the horizontal speed doubles the range. Increasing the height increases the time of flight and therefore the range, but as a square-root relationship: quadrupling the height only doubles the range.

This leads to a counterintuitive result: a bullet fired horizontally from a rifle and a bullet simply dropped from the same height will hit the ground at exactly the same time — their vertical motions are identical. The fired bullet travels much further horizontally, but takes the same time to fall.

What is the effect of launch angle on range (for angled projectiles)?

When a projectile is launched at an angle (rather than horizontally), the initial velocity has both horizontal and vertical components:

• Horizontal component: v_x = v × cos(θ)
• Vertical component (upward): v_y = v × sin(θ)

The projectile rises until its vertical velocity reaches zero, then falls back. The time of flight is longer for steeper angles (the object goes higher and takes longer to come down). For a given initial speed, the range is maximum at a launch angle of 45°, where the best compromise between horizontal speed and time of flight is achieved. This result is on the boundary of GCSE and A-level content and may appear in the highest mark GCSE questions.

Frequently asked questions

Why does the time of flight not depend on horizontal speed?

The time of flight is entirely determined by the vertical motion — specifically, how long it takes gravity to bring the object from its launch height down to the ground. Gravity acts only vertically, so it affects only the vertical motion. The horizontal speed has no effect on gravity or on the rate of vertical fall. This is a direct consequence of the independence of horizontal and vertical components: however fast the ball is thrown horizontally, it falls at exactly the same rate as if it had simply been dropped.

How do you find the actual speed of a projectile at any moment?

At any moment, the projectile has a horizontal component of velocity (v_x = constant) and a vertical component (v_y = g × t, increasing downward). The actual speed is the vector sum of these two components. Using Pythagoras's theorem: actual speed = √(v_x² + v_y²). The direction of motion is at an angle to the horizontal given by: tan(θ) = v_y ÷ v_x. At launch (t = 0), the speed equals the horizontal speed (v_y = 0). At landing, the speed is greatest because v_y has been building throughout the fall.

What is the shape of a projectile's path?

The path (trajectory) of a projectile in the absence of air resistance is a parabola. This can be shown mathematically: horizontal distance x = v_x × t, so t = x ÷ v_x. Vertical distance y = ½ g t² = ½ g (x ÷ v_x)². Because y is proportional to x² (with a constant coefficient), the path is a parabola. In reality, air resistance causes the trajectory to deviate from a perfect parabola, making the real range shorter and the descent steeper than predicted.

Why can you use SUVAT equations for the vertical motion of a projectile?

The SUVAT equations (s = ut + ½at², v = u + at, v² = u² + 2as, etc.) apply whenever an object undergoes constant acceleration. In projectile motion, gravity provides a constant downward acceleration of g = 10 m/s² (assuming the projectile does not travel so high that g changes significantly). The vertical motion therefore satisfies the conditions for SUVAT. The horizontal motion has zero acceleration, so it only needs the simpler equation: horizontal distance = horizontal velocity × time.

For predict-first GCSE physics with Professor Newton — decomposing motion into components and predicting the landing point before any calculation — visit aitutors.me.