Physics • Year 12 • Module 5 • Lesson 4
Phase 1 Consolidation
Apply projectile motion formulae to numerical problems, error identification, and real-data scenarios from Band 3 to Band 5.
1. Numerical problems — Band 3–4
Show all working. Include formula, substitution, and final answer with units. 12 marks
1.1 A cricket ball is thrown at 18 m s−1, 35° above horizontal. Find: (a) the horizontal and vertical components of the initial velocity, (b) the total flight time on level ground, and (c) the horizontal range. 4 marks
1.2 A ball rolls off a desk 0.85 m high with a horizontal velocity of 3.2 m s−1. Find: (a) the time the ball takes to reach the floor, and (b) how far from the base of the desk it lands. 4 marks
1.3 A rugby ball is kicked at 28 m s−1, 40° above horizontal on level ground. Find the maximum height it reaches above the kick point. 2 marks
1.4 A stone is thrown horizontally at 12 m s−1 from a cliff 45 m above the ground. Find the horizontal distance from the base of the cliff where the stone lands. 2 marks
2. Error identification — Band 4
Each item below shows a student’s working. Identify the error and write the corrected working. 8 marks (2 per error: 1 identify, 1 correct)
2.1 Problem: A ball is launched at 20 m s−1, 50°. Find the time to reach maximum height.
Student answer: “\(t = v/g = 20/9.8 = 2.04\) s”
What is the error?
Corrected working:
2.2 Problem: A ball is launched at 15 m s−1, 45° from the edge of a 20 m cliff. Find the horizontal range to where it lands.
Student answer: “\(R = \frac{15^2\sin(90°)}{9.8} = 22.96\) m”
What is the error?
Corrected working (outline the correct method):
2.3 Problem: A stone is thrown upward at 14 m s−1. Find its displacement after 2.0 s.
Student answer: “\(s_y = 14(2.0) + \frac{1}{2}(9.8)(2.0)^2 = 28 + 19.6 = 47.6\) m”
What is the error?
Corrected working:
2.4 Problem: A ball is at maximum height. A student states: “The velocity of the ball is zero at this point, so it momentarily stops completely.”
What is the error?
Corrected statement:
3. Data interpretation — projectile comparison table
Two projectiles, P and Q, are both launched at 20 m s−1 on level ground. P is launched at 30°; Q at 60°. Complete the table below by calculating each quantity. Show your working in the spaces provided below. 8 marks (1 each)
| Quantity | Projectile P (30°) | Projectile Q (60°) |
|---|---|---|
| Initial horizontal component \(v_x\) (m s−1) | ||
| Initial vertical component \(v_{y0}\) (m s−1) | ||
| Total flight time \(T\) (s) | ||
| Maximum height \(h_{\max}\) (m) | ||
| Horizontal range \(R\) (m) |
Working space:
3.1 Based on your completed table, state two quantities that are the same for P and Q, and two that differ. Explain why range is the same using the range formula. 2 marks
Q1.1 — Cricket ball throw (4 marks)
(a) \(v_x = 18\cos35° = \mathbf{14.74 \approx 14.7 \text{ m s}^{-1}}\); \(v_{y0} = 18\sin35° = \mathbf{10.32 \approx 10.3 \text{ m s}^{-1}}\) (1 mark)
(b) \(T = 2v_{y0}/g = 2 \times 10.32/9.8 = \mathbf{2.11 \text{ s}}\) (1 mark)
(c) \(R = v_x \times T = 14.74 \times 2.11 = \mathbf{31.1 \text{ m}}\) (1 mark)
Or: \(R = 18^2\sin70°/9.8 = 31.1\) m. (Award 1 mark for consistent method)
Marking note. Award 1 mark for each part. ECF (follow-through) applies: if (a) is wrong but (b)/(c) follow correctly from (a), award method marks.
Q1.2 — Ball off desk (4 marks)
(a) Horizontal launch, so \(v_{y0} = 0\). \(s_y = \tfrac{1}{2}gt^2 \Rightarrow 0.85 = \tfrac{1}{2}(9.8)t^2\) → \(t = \sqrt{2 \times 0.85/9.8} = \mathbf{0.416 \approx 0.42 \text{ s}}\) (2 marks: 1 formula, 1 answer)
(b) \(s_x = v_x \cdot t = 3.2 \times 0.416 = \mathbf{1.33 \approx 1.33 \text{ m}}\) (2 marks)
Marking note. Using \(g = 9.81\) is acceptable (\(t \approx 0.416\) s, \(s_x \approx 1.33\) m). ECF applies.
Q1.3 — Rugby ball max height (2 marks)
\(v_{y0} = 28\sin40° = 18.00\) m s−1 (1 mark)
\(h_{\max} = v_{y0}^2/(2g) = 18.00^2/(2 \times 9.8) = 324/19.6 = \mathbf{16.5 \text{ m}}\) (1 mark)
Q1.4 — Horizontal cliff launch (2 marks)
\(t = \sqrt{2s_y/g} = \sqrt{2 \times 45/9.8} = \sqrt{9.18} = 3.03\) s (1 mark)
\(s_x = v_x \cdot t = 12 \times 3.03 = \mathbf{36.4 \text{ m}}\) (1 mark)
Q2.1 — Error: using total speed instead of vertical component (2 marks)
Error: The student used total speed \(v = 20\) m s−1 instead of the vertical component \(v_{y0} = v\sin\theta\) [1].
Corrected: \(v_{y0} = 20\sin50° = 15.32\) m s−1. At max height, \(v_y = 0\): \(t = v_{y0}/g = 15.32/9.8 = \mathbf{1.56 \text{ s}}\) [1].
Q2.2 — Error: using range formula from a cliff (2 marks)
Error: The range formula \(R = v^2\sin(2\theta)/g\) only applies when launch and landing heights are equal. The projectile is launched from a 20 m cliff, so the landing height is lower than the launch height [1].
Corrected method: Resolve components: \(v_x = 15\cos45° = 10.61\) m s−1; \(v_{y0} = 10.61\) m s−1. Use \(s_y = v_{y0}t + \tfrac{1}{2}(-9.8)t^2\) with \(s_y = -20\) m to find \(t \approx 3.05\) s, then \(s_x = v_x t = 10.61 \times 3.05 \approx 32.4\) m [1].
Q2.3 — Error: positive gravity sign (2 marks)
Error: The student used \(+9.8\) m s−2 for gravity. Since the ball is thrown upward and up is positive, acceleration due to gravity must be \(-9.8\) m s−2 [1].
Corrected: \(s_y = 14(2.0) + \tfrac{1}{2}(-9.8)(2.0)^2 = 28 - 19.6 = \mathbf{+8.4 \text{ m}}\). The ball is 8.4 m above the launch point after 2.0 s [1].
Q2.4 — Error: velocity = 0 at max height (2 marks)
Error: The student confused the vertical component of velocity being zero with the total velocity being zero [1].
Corrected statement: At maximum height, only the vertical component \(v_y = 0\). The horizontal component \(v_x = v\cos\theta\) remains constant throughout the flight (no horizontal force). The ball is still moving horizontally at maximum height with speed equal to \(v_x\) [1].
Q3 — Comparison table & explanation
Projectile P (30°): \(v_x = 20\cos30° = 17.3\) m s−1; \(v_{y0} = 20\sin30° = 10.0\) m s−1; \(T = 2 \times 10.0/9.8 = 2.04\) s; \(h_{\max} = 10.0^2/19.6 = 5.10\) m; \(R = 20^2\sin60°/9.8 = 35.3\) m
Projectile Q (60°): \(v_x = 20\cos60° = 10.0\) m s−1; \(v_{y0} = 20\sin60° = 17.3\) m s−1; \(T = 2 \times 17.3/9.8 = 3.53\) s; \(h_{\max} = 17.3^2/19.6 = 15.3\) m; \(R = 20^2\sin120°/9.8 = 35.3\) m
3.1: Same: horizontal range (35.3 m each); the range formula gives \(R = v^2\sin(2\theta)/g\) and \(\sin(2 \times 30°) = \sin60° = \sin120° = \sin(2 \times 60°)\) (complementary angles). Different: maximum height (5.1 m vs 15.3 m) and total flight time (2.04 s vs 3.53 s).
Marking note. Award 1 mark per cell in the table (8 marks total across column pairs). ECF applies if components are computed correctly. 2 marks for Q3.1: 1 for identifying same/different quantities, 1 for the \(\sin\) identity explanation.