Explain why the horizontal range of a projectile launched at 30° above the horizontal is the same as the range of a projectile launched at 60° above the horizontal, assuming both are launched with the same initial speed from the same level ground and air resistance is negligible.
Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).
Model answer (4 marks)
The horizontal range R is given by R=v₀cosθ·t_f, where t_f is the time of flight. For a projectile launched from level ground, t_f=2v₀sinθ/g. Thus R=v₀²sin2θ/g. The angles 30° and 60° are complementary, so sin2θ is the same for both (sin60°=sin120°). Therefore R is identical for 30° and 60°. In other words, at 30° the horizontal component v₀cosθ is larger but the flight time 2v₀sinθ/g is shorter; at 60° the horizontal component is smaller but the flight time is longer. The two effects cancel, giving the same range.
Examiner tips
- Show the formula R=v₀²sin2θ/g and note sin2θ is equal for complementary angles.
- Explain the trade‑off between horizontal speed and flight time.
- Use the fact that 30°+60°=90° to justify equality.
- Keep the answer concise and use correct symbols.
Mark scheme (4 marks)
- The horizontal range depends on both the horizontal component of velocity and the time of flight.
- At 30°, the horizontal component is larger but the time of flight is shorter; at 60°, the horizontal component is smaller but the time of flight is longer.
- The time of flight is determined by the vertical component of initial velocity, which is larger at 60°, giving a proportionally longer flight time.
- 30° and 60° are complementary angles (sum to 90°), so the product of the horizontal component and the time of flight is identical for both, yielding equal ranges.
Key terms in this question
Related
- All IB DP Physics Standard Level (2023 syllabus) revision notes →
- How to answer a "Explain" question →
- Decode the mark scheme abbreviations →
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