Class 11 Physics Chapter 3: Motion in a Plane case study question with answers
One case-based question of the kind CBSE-pattern school exams ask: a passage and a table, then 3 parts worth 4 marks in all, with the answer and the marking scheme for each. Written from the NCERT chapter.
| angle of projection | 15° | 30° | 45° | 60° | 75° |
|---|---|---|---|---|---|
| range (m) | 20.0 | 34.6 | 40.0 | 34.6 | 20.0 |
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Answer: (b) 5 m
hm = (v₀ sin θ₀)²/2g = (20 × 0.5)²/20 = 5 m. 15 m is the height for the 60° throw.
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Answer: (d) 3.46 s
Tf = 2v₀ sin θ₀/g = 2 × 20 × (√3/2)/10 = 2√3 ≈ 3.46 s. 1.73 s is the time to the top only.
Answer and marking scheme
Same: the range, 34.6 m, since sin 60° = sin 120°. Different: the maximum height, 5 m at 30° and 15 m at 60°; the time of flight, 2 × 20 × 0.5/10 = 2 s at 30° and 3.46 s at 60°.
- range 34.6 m the same for both (1)
- heights 5 m and 15 m, or times 2 s and 3.46 s, as the differences (1)
Answer and marking scheme
R ∝ v₀²: halving the speed divides the range by 4, so 40.0/4 = 10 m. hm ∝ v₀² too: at 20 m s⁻¹ and 45°, hm = (20 × 0.707)²/20 = 10 m, so it falls to 10/4 = 2.5 m.
- range ∝ v₀², new range 10 m (1)
- maximum height falls by a factor 4, from 10 m to 2.5 m (1)
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