A-Level · A-Level Physics
A-Level Physics projectile motion
What is projectile motion?
Projectile motion describes the path of an object moving under gravity alone after launch. The motion decomposes into independent horizontal (constant velocity) and vertical (constant downward acceleration g) components. Resolve the initial velocity into v cosθ (horizontal) and v sinθ (vertical), then apply suvat to each component independently. Range, maximum height and time of flight are the three exam-favourite calculations.
Worked example
A projectile is launched at 30 m/s at an angle of 45° above the horizontal from ground level. Find the range (assume g = 9.81 m/s²).
- Resolve initial velocity: u_x = 30 cos(45°) = 21.2 m/s; u_y = 30 sin(45°) = 21.2 m/s.
- Find time of flight using vertical motion. The projectile returns to ground level, so vertical displacement = 0. Use s = u_y · t − (1/2)g · t² = 0 → t(u_y − (1/2)g · t) = 0 → t = 2u_y / g = 2(21.2) / 9.81 = 4.32 s.
- Find horizontal range: R = u_x · t = 21.2 × 4.32 = 91.6 m.
Decomposing the motion
The key insight: horizontal and vertical motion are independent. Horizontal velocity stays constant (no air resistance assumed); vertical velocity changes by g per second. Resolve the initial velocity v at angle θ into:
- Horizontal component: u_x = v cosθ.
- Vertical component: u_y = v sinθ (upward positive).
- Apply suvat separately to each component.
Key formulas
For a projectile launched from ground level at angle θ with speed v:
- Time of flight: t = 2v sinθ / g.
- Maximum height: H = (v sinθ)² / 2g.
- Range: R = v² sin(2θ) / g.
- Range is maximised at θ = 45° (when launch and landing heights are equal).
Where it appears in the exam
AQA 7408, Edexcel 9PH0 and OCR A H556 examine projectile motion on the Mechanics paper. Common variations: launch from a height (not ground level), air resistance considered qualitatively, finding the angle of impact, projectile passing through a specified point.
Common mistakes
- Forgetting to resolve the initial velocity into components.
- Sign errors on the vertical component — choose 'up positive' or 'down positive' and stick with it.
- Confusing time-of-flight (full trajectory) with time-to-max-height (half the trajectory, only when launch and landing heights are equal).
- Using suvat with mixed-axis quantities — apply each suvat equation to one axis at a time.
Frequently asked
- Why is the range maximised at 45°?
- When launch and landing heights are equal, R = v² sin(2θ) / g. The maximum of sin(2θ) is at 2θ = 90°, so θ = 45°. Note: this only holds when the launch and landing heights are the same. From a cliff (different heights), the optimal angle is less than 45°.
- Do A-Level questions include air resistance?
- Quantitative projectile motion problems at A-Level assume air resistance is negligible. Air resistance is discussed qualitatively (shorter range, asymmetric trajectory) but not calculated in projectile-motion questions.
- How do I find the velocity at any point in the trajectory?
- Use suvat on each component: v_x = u_x (constant); v_y = u_y − g · t. The total speed is √(v_x² + v_y²), and the angle below horizontal is arctan(v_y / v_x).
A-Level Physics glossary terms
- Projectile motionProjectile motion describes the path of an object moving under gravity alone after being launched at an angle. Horizontal motion has constant velocity; vertical motion has constant acceleration g downwards. Resolve initial velocity into horizontal (v cosθ) and vertical (v sinθ) components, then apply suvat to each component independently. Range, maximum height and time of flight are the three exam-favourite calculations.
- Hooke's LawHooke's Law states that the force required to extend (or compress) a spring is proportional to the extension, provided the spring is within its elastic limit: F = kx, where k is the spring constant (N/m) and x is the extension from natural length. Past the elastic limit the relationship becomes non-linear and the spring may deform permanently. A force-extension graph rewards correct gradient calculation (gradient = k).
- Simple harmonic motionSimple harmonic motion describes oscillation where the restoring force is proportional to displacement and directed toward equilibrium: a = −ω²x. The system oscillates with constant period T = 2π/ω regardless of amplitude. SHM examples: mass on a spring (ω = √(k/m)), simple pendulum (ω = √(g/L) for small angles), molecular vibrations. A-Level Physics examines displacement, velocity and acceleration equations, energy in SHM and damping.
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