QCE Physics — Unit 3
Projectile Motion — Flashcards & Quiz
Projectile motion combines two independent one-dimensional motions: constant horizontal velocity and constant vertical acceleration (g = 9.8 m s⁻² downward). QCE Physics Unit 3 treats projectiles as a go-to calculation and derivation topic, and you are expected to resolve the launch velocity into components, apply kinematic equations separately, and solve for range, maximum height or time of flight.
Key Points
- Horizontal motion: constant velocity uₓ = u cosθ; no acceleration (ignore air resistance).
- Vertical motion: constant acceleration a = –g; initial velocity uᵧ = u sinθ; use SUVAT equations.
- Time of flight (level ground): t = 2u sinθ / g; derived from setting vertical displacement to zero.
- Range: R = u² sin(2θ) / g; maximum range is at θ = 45° on level ground with no air resistance.
- Maximum height: H = (u sinθ)² / (2g); occurs when vertical velocity = 0.
- At the top of the trajectory, vertical velocity is zero but horizontal velocity is unchanged — velocity is purely horizontal there.
Common Mistakes to Avoid
- Adding horizontal and vertical velocities as scalars — they are perpendicular components.
- Using sinθ for horizontal component — it's cosθ (when θ is measured from horizontal).
- Forgetting that horizontal velocity is constant throughout the motion.
- Applying the range formula to uneven ground or with air resistance — it only works on level ground with no air resistance.
- Mixing up signs: if up is positive then g = –9.8 m s⁻², not +9.8.
Exam Strategy
QCAA Unit 3 projectile questions give an initial speed and angle, then ask for time, range, maximum height, or the velocity at a specific time. Method: (1) resolve u into uₓ = u cosθ and uᵧ = u sinθ, (2) treat horizontal and vertical independently using SUVAT, (3) use time as the link between them, (4) state any assumptions (level ground, no air resistance).
Sample Flashcards
Q1: Describe the two independent components of projectile motion.
Horizontal: constant velocity (a = 0), x = v_x t. Vertical: constant acceleration g = 9.8 m s⁻² downward. Components are independent — horizontal does not affect vertical.
Q2: Find range and max height for a projectile at angle θ.
v_x = v cos θ, v_y = v sin θ. H = v_y²/(2g). T = 2v_y/g. R = v² sin 2θ / g. Max range at 45°.
Q3: How do you find velocity at any point during flight?
v = √(v_x² + v_y²), v_y = v_y₀ − gt. Direction: tan θ = |v_y|/v_x below horizontal.
Q4: How does air resistance affect projectiles?
Drag reduces horizontal speed, max height and range. Trajectory becomes asymmetric — steeper descent than ascent.
Sample Quiz Questions
Q1: Horizontal velocity stays constant without air resistance.
Answer: TRUE
No horizontal force → no horizontal acceleration.
Q2: Launch angles 30° and 60° give the same range at equal speed.
Answer: TRUE
Complementary angles: sin 2(30°) = sin 2(60°).
Q3: At the peak, both velocity and acceleration are zero.
Answer: FALSE
Vertical velocity = 0 but horizontal velocity persists and g acts throughout.
Revision Tip
Projectile problems all follow the same decomposition pattern — drill a Revizi flashcard deck with 8+ scenarios (horizontal launch, angled launch, from a cliff) to lock in the method.
Last updated: March 2026 · 4 flashcards · 3 quiz questions