OCR Specification focus:
‘Model two-dimensional projectile motion with constant velocity in one direction and constant acceleration perpendicular to it.’
Projectile motion is a key application of kinematics that combines uniform motion horizontally with uniformly accelerated motion vertically. Understanding these independent components allows precise prediction of an object’s path through space.
The Nature of Projectile Motion
Projectile motion occurs when an object is launched into the air and moves under the influence of gravity alone, assuming negligible air resistance. This means the only force acting on the object after launch is its weight, directed vertically downwards. The motion can therefore be divided into two independent components — horizontal and vertical — that together determine the overall trajectory.
Independent Components
The horizontal and vertical components of projectile motion are treated separately:
Horizontal motion involves constant velocity, since there is no acceleration in this direction.
Vertical motion involves constant acceleration, due to the effect of gravity (g).
Each component follows the equations of motion independently, and the results are then combined to describe the full two-dimensional path.
Figure 3.35 illustrates projectile motion analysed as two independent one-dimensional motions. Panel (b) shows constant horizontal velocity (aₓ = 0); panel (c) shows vertical motion under constant acceleration (aᵧ = −g); panel (d) recombines the components to give the resultant velocity vector at any point. Source.
Horizontal Motion: Constant Velocity
Because air resistance is ignored, the horizontal component of velocity remains constant throughout the flight.
Constant velocity: Motion in which both the speed and direction remain unchanged, implying zero acceleration.
The distance travelled horizontally depends solely on the time the projectile is in the air. Since time is common to both components, it provides the crucial link between them.
EQUATION
—-----------------------------------------------------------------
Horizontal displacement (sₓ) = uₓ × t
uₓ = Horizontal component of initial velocity (m s⁻¹)
t = Time of flight (s)
—-----------------------------------------------------------------
This simple relationship allows determination of how far a projectile travels along the horizontal plane while gravity influences its vertical motion.
Vertical Motion: Constant Acceleration
The vertical component of motion behaves as a case of uniformly accelerated motion, with a constant downward acceleration equal to g, the acceleration due to gravity. For most calculations, g ≈ 9.81 m s⁻² near Earth’s surface.
Constant acceleration: A situation where velocity changes by equal amounts in equal time intervals, producing uniform change in motion.
The vertical velocity and displacement change continually under gravity, following the standard equations of motion for constant acceleration.
EQUATION
—-----------------------------------------------------------------
vᵧ = uᵧ + a t
vᵧ = Final vertical velocity (m s⁻¹)
uᵧ = Initial vertical velocity (m s⁻¹)
a = Acceleration (here g = 9.81 m s⁻², downward)
t = Time (s)
—-----------------------------------------------------------------
EQUATION
—-----------------------------------------------------------------
sᵧ = uᵧ t + ½ a t²
sᵧ = Vertical displacement (m)
uᵧ = Initial vertical velocity (m s⁻¹)
a = Acceleration due to gravity (m s⁻²)
t = Time (s)
—-----------------------------------------------------------------
Between these two relationships, students can find the vertical position and velocity of a projectile at any instant.
The time of flight, maximum height, and overall path depend entirely on this vertical motion because it determines when the projectile returns to its original height.
The Independence of Components
A key concept in this subsubtopic is the independence of horizontal and vertical motion. Each component acts as if the other does not exist:
Horizontal velocity has no effect on vertical acceleration.
Vertical acceleration (gravity) has no effect on horizontal velocity.
This independence means a projectile launched horizontally from a height and a body simply dropped from the same height will hit the ground simultaneously, assuming equal vertical starting positions and negligible air resistance.
Combining the Components
Although the components are analysed separately, they can be recombined vectorially to find the projectile’s resultant velocity at any point.
EQUATION
—-----------------------------------------------------------------
Resultant velocity (v) = √(vₓ² + vᵧ²)
vₓ = Horizontal velocity (constant, m s⁻¹)
vᵧ = Vertical velocity at that instant (m s⁻¹)
—-----------------------------------------------------------------
The direction of motion (θ) relative to the horizontal can be found using the trigonometric relationship:
θ = tan⁻¹(vᵧ / vₓ)
This enables full description of the projectile’s instantaneous motion in both magnitude and direction.
Understanding the Trajectory
The overall trajectory of a projectile is a parabola. This shape results from combining:
Constant horizontal velocity (linear displacement over time)
Constant vertical acceleration (quadratic displacement over time)
Parabolic trajectory of a projectile with initial speed u at angle θ, showing range (x) and maximum height (y). The shape arises from constant horizontal velocity combined with constant vertical acceleration. Labels are minimal and suitable for A-Level revision. Source.
Because the vertical displacement follows a squared time dependence (½ g t²), the curve steepens as the object ascends and descends. The top of the parabola corresponds to the instant when vertical velocity equals zero, marking the maximum height.
Key Points About the Trajectory
The path is symmetrical for projectiles launched and landing at the same vertical level.
Time of ascent equals time of descent (ignoring air resistance).
At the maximum height, vertical velocity is zero, but horizontal velocity remains unchanged.
The range — the total horizontal distance travelled — depends on both initial speed and launch angle.
Practical Relevance and Modelling
The OCR specification highlights the need to model projectile motion using these independent components. This model assumes:
No air resistance (idealised conditions)
Constant acceleration due to gravity
Flat, uniform terrain
Point-mass projectile (size and rotation ignored)
These simplifications allow straightforward mathematical analysis using SUVAT equations and trigonometric relationships.
In practical applications such as sports, ballistics, or engineering, small deviations from these assumptions (like air drag or wind) can significantly alter trajectories. Nevertheless, this ideal model forms the foundation for understanding more complex motion in real situations.
Summary of Conceptual Framework
Students must be confident in:
Resolving motion into horizontal and vertical components using trigonometric methods.
Applying constant velocity equations horizontally and constant acceleration equations vertically.
Interpreting how these independent motions combine to form a parabolic trajectory.
Recognising that time is the shared variable linking both components.
This understanding underpins the analysis of all projectile motion problems within A-Level Physics and provides essential groundwork for experimental and mathematical treatment of two-dimensional motion.
FAQ
The time a projectile remains in the air depends entirely on its vertical motion. Specifically, it’s determined by the initial vertical velocity and the acceleration due to gravity.
A projectile launched at a greater angle (up to 45°) will have a larger vertical component of velocity, keeping it in the air longer. Conversely, a purely horizontal launch has zero vertical velocity initially, and the time of flight depends only on the height from which it’s released.
Ignoring air resistance, the formula used is:
Time of flight = (2uᵧ) / g, for projectiles launched and landing at the same height.
A projectile’s path is parabolic because its horizontal and vertical motions behave differently:
The horizontal velocity remains constant.
The vertical velocity changes uniformly due to gravity.
When these two motions are combined, the result is a quadratic relationship between vertical displacement and horizontal displacement, producing a smooth parabola.
This parabolic shape is symmetrical when the projectile lands at the same level from which it was launched.
The range depends on both the magnitude and angle of the initial velocity.
A small angle produces a short, flat trajectory.
A large angle gives more height but less horizontal distance.
The maximum range (on level ground, without air resistance) occurs at a 45° launch angle.
For complementary angles (e.g. 30° and 60°) with the same launch speed, the ranges are equal but the trajectories differ in height and time of flight.
At the highest point:
The vertical velocity is zero — the projectile momentarily stops rising before it begins to fall.
The horizontal velocity remains unchanged, as there is no horizontal acceleration.
The acceleration due to gravity still acts downward at this point, ensuring that the projectile immediately begins to accelerate downward again after reaching the peak.
Air resistance introduces a horizontal deceleration and reduces vertical velocity during both ascent and descent.
The path is no longer perfectly parabolic.
The range and maximum height both decrease.
The time of flight can increase or decrease depending on conditions, but total distance travelled is always less than predicted by the ideal model.
This is why real-world trajectories, like those of a ball or arrow, fall short of the theoretical predictions made without air resistance.
Practice Questions
Question 1 (2 marks)
A ball is projected horizontally from the top of a 20 m high cliff with a constant horizontal velocity of 6.0 m s⁻¹.
(a) State and explain what happens to the horizontal velocity of the ball during its flight.
Mark scheme:
1 mark: States that the horizontal velocity remains constant.
1 mark: Explains that there is no horizontal acceleration because air resistance is neglected (no horizontal forces act).
Question 2 (5 marks)
A stone is thrown with an initial velocity of 15 m s⁻¹ at an angle of 40° above the horizontal. The acceleration due to gravity is 9.81 m s⁻².
(a) Calculate the time of flight of the stone.
(b) Explain, using the independence of horizontal and vertical motion, why the horizontal component of the velocity remains unchanged during the flight while the vertical component changes continuously.
Mark scheme:
(a) Calculation (3 marks)
1 mark: Resolves velocity into vertical component uᵧ = 15 sin(40°).
1 mark: Recognises that time of flight T = (2uᵧ) / g.
1 mark: Substitutes correctly and obtains T ≈ 1.96 s (accept 1.9–2.0 s range).
(b) Explanation (2 marks)
1 mark: States that horizontal and vertical components are independent; gravity acts only vertically.
1 mark: Therefore, horizontal velocity remains constant while vertical velocity changes due to constant vertical acceleration (g).
