Analyse the flight of any projectile by treating horizontal and vertical motion independently using the suvat equations.
AQA A-Level Physics · Unit 4: Mechanics
↔️Independence of motion Explain why horizontal and vertical components are treated separately
📐Initial components Resolve initial velocity into horizontal and vertical components
📏suvat equations Apply the five kinematic equations to projectile problems
⏱️Time of flight Calculate total time in the air for a symmetric projectile
📈Maximum height Find the maximum height reached by a projectile
🎯Range Calculate horizontal range and identify optimum launch angle
The Key Principle: Independence of Motion
A projectile is any object that moves under the influence of gravity alone (air resistance neglected). The crucial insight — first realised by Galileo — is that horizontal and vertical motions are completely independent of each other.
Horizontal: No force acts horizontally (assuming no air resistance), so horizontal velocity is constant throughout the flight. This is uniform motion.
Vertical: Gravity acts downward with g = 9.81 m s⁻². The vertical motion is uniformly accelerated (or decelerated on the way up).
The horizontal and vertical components of a projectile's motion are independent. Treat them as two separate suvat problems that share only one variable: time.
This independence can be demonstrated experimentally: two balls released simultaneously — one dropped vertically, one projected horizontally — hit the ground at exactly the same time. The horizontal velocity of the projectile does not affect the time taken to fall vertically.
Projectile: An object given an initial velocity and then moving freely under gravity only.
Setting Up a Projectile Problem
For a projectile launched with speed u at angle θ above the horizontal:
Horizontal: uₓ = u cosθ aₓ = 0
Vertical: uᵧ = u sinθ aᵧ = −g = −9.81 m s⁻²
The sign convention matters: take upward as positive and rightward as positive. Gravity therefore has a = −9.81 m s⁻².
The five suvat equations (for constant acceleration) are:
v = u + at
s = ut + ½at²
v² = u² + 2as
s = ½(u + v)t
s = vt − ½at²
Symbol
Quantity
Unit
s
Displacement
m
u
Initial velocity
m s⁻¹
v
Final velocity
m s⁻¹
a
Acceleration
m s⁻²
t
Time
s
Apply these equations separately to the horizontal (with a = 0) and vertical directions. Time t is the linking variable between the two.
Key Results: Range, Height, Time of Flight
For a projectile launched at speed u, angle θ, from ground level on flat ground:
Time of flight: T = 2u sinθ / g
Maximum height: H = u² sin²θ / (2g)
Horizontal range: R = u² sin(2θ) / g
These are derived results — you should be able to derive them from suvat rather than just memorise them. However, knowing them saves time in exams.
Key observations:
Maximum range occurs at θ = 45° (since sin 2θ is maximised when 2θ = 90°)
Angles that give equal ranges: θ and (90° − θ). For example, 30° and 60° give the same range.
At maximum height, the vertical velocity is zero (vᵧ = 0) — the projectile moves horizontally only.
These range and height formulas only apply to projectiles launched and landing at the same height. For cliffs, towers, or inclined ground, you must work from suvat directly.
The trajectory of a projectile (y as a function of x) is a parabola. You can derive this by eliminating t between the horizontal and vertical displacement equations.
Velocity at Any Point
At any time t during the flight, the projectile has:
vₓ = u cosθ (constant)
vᵧ = u sinθ − gt
The resultant speed at that moment: v = √(vₓ² + vᵧ²)
The direction of motion: α = tan⁻¹(vᵧ / vₓ) below the horizontal (when vᵧ is negative, i.e. the object is descending).
Note that at maximum height, vᵧ = 0, so the resultant velocity equals vₓ = u cosθ (purely horizontal).
For a symmetric flight (landing at the same height as launch), the speed at landing equals the launch speed u, but the vertical component is now downward: vᵧ = −u sinθ. The angle below horizontal equals the launch angle above horizontal.
The projectile's speed is minimum at maximum height (= u cosθ) and equal to the launch speed at landing.
When projectile problems involve a cliff or elevated platform, the vertical displacement s at landing will be negative (below the launch point). Substituting the correct (negative) value into suvat automatically accounts for this.
A ball is kicked at 20 m s⁻¹ at 35° above the horizontal. Calculate (a) the time of flight and (b) the horizontal range. (Assume g = 9.81 m s⁻², flat ground.)
1Resolve: uₓ = 20 cos35° = 16.38 m s⁻¹; uᵧ = 20 sin35° = 11.47 m s⁻¹
2Vertical, find time of flight: use s = uᵧt + ½at² with s = 0, a = −9.81: 0 = 11.47t − 4.905t²
3t(11.47 − 4.905t) = 0 → t = 0 (launch) or t = 11.47/4.905 = 2.34 s
4Horizontal: R = uₓ × t = 16.38 × 2.34 = 38.3 m
Time of flight = 2.34 s; Range = 38.3 m
A stone is thrown horizontally at 12 m s⁻¹ from the top of a cliff 45 m high. How far from the base of the cliff does it land?
1Horizontal: uₓ = 12 m s⁻¹, aₓ = 0
2Vertical: uᵧ = 0 (thrown horizontally), s = −45 m (downward), a = −9.81 m s⁻²
4Speed at max height = uₓ = 25 cos50° = 16.07 m s⁻¹
Maximum height = 18.7 m; Speed at maximum height = 16.1 m s⁻¹
Q1. A projectile is launched horizontally. Which statement about its motion is correct?
Q2. At what launch angle is the horizontal range of a projectile maximised (on flat ground)?
Q3. A ball is dropped from rest from height h. A second ball is projected horizontally at the same moment from the same height. Which hits the ground first?
Q4. A ball is launched at 15 m s⁻¹ horizontally from a table 1.25 m high. How long does it take to reach the ground? (g = 9.81 m s⁻²)
Q5. A projectile is launched at 30 m s⁻¹ at an angle of 40° above horizontal. What is the vertical component of velocity at the highest point of the trajectory?
Challenge Q1. A cannon on top of a 60 m cliff fires a shell at 80 m s⁻¹ at 25° above horizontal. Find the total horizontal distance from the cliff base where the shell lands.
Challenge Q2. Show that two launch angles θ and (90° − θ) give the same horizontal range on flat ground. Start from R = u²sin(2θ)/g.
Challenge Q3. A footballer kicks a ball at 22 m s⁻¹ aiming at a goal 28 m away. The crossbar is 2.44 m high. At what angle(s) should the ball be kicked so it just clears the crossbar? (Derive the trajectory equation and solve numerically.)