Physics • Year 12 • Module 5 • Lesson 4
Phase 1 Consolidation
Lock in the six projectile motion formulae, their conditions of use, and the six common errors before tackling harder questions.
1. Formula–condition match
Draw a line (or write the letter) from each formula in Column A to its correct condition of use in Column B. 6 marks (1 each)
| Column A — Formula | Column B — Condition of use |
|---|---|
| A. \(x = v_x \cdot t\) | i. Level ground only; launch height equals landing height. |
| B. \(y = v_{y0}t + \tfrac{1}{2}gt^2\) | ii. Finding horizontal displacement; always valid (no height restriction). |
| C. \(v_y = v_{y0} + gt\) | iii. Finding vertical velocity at any instant; set \(v_y = 0\) for maximum height time. |
| D. \(v_y^2 = v_{y0}^2 + 2gy\) | iv. Finding maximum height or linking vertical velocity to displacement without time. |
| E. \(R = \dfrac{v^2\sin(2\theta)}{g}\) | v. Finding vertical displacement at a given time; use for cliff launches (solve quadratic). |
| F. \(T = \dfrac{2v_{y0}}{g}\) | vi. Total flight time on level ground only; time up equals time down. |
Write your answers here: A–____ B–____ C–____ D–____ E–____ F–____
2. True or false — with correction
Circle T or F for each statement. If the statement is false, write the corrected version on the line below it. 12 marks (1 T/F + 1 correction each)
2.1 The horizontal component of a projectile’s velocity is constant throughout its flight because no horizontal force acts (ignoring air resistance). T / F
2.2 The range formula \(R = v^2\sin(2\theta)/g\) can be used when a projectile is launched from the top of a cliff. T / F
2.3 At maximum height, the vertical component of velocity equals zero but the horizontal component remains unchanged. T / F
2.4 A projectile launched at 30° and one launched at 60° (same speed, level ground) have the same maximum height. T / F
2.5 Flight time is determined entirely by the vertical motion and is independent of horizontal velocity. T / F
2.6 When taking upward as positive, the acceleration due to gravity must be entered as \(+9.8\) m s−2 in vertical motion equations. T / F
3. Fill-in-the-blank paragraph
Use the word bank to complete the passage. Each word is used once. 8 marks (1 per blank)
Word bank:
component · constant · equal · gravity · height · parabolic · vertical · zero
In projectile motion, the only force acting is ___________, which acts downward. The horizontal velocity remains ___________ throughout flight because no horizontal force acts. The vertical velocity changes at a rate of 9.8 m s−2 downward due to gravity. A projectile follows a ___________ path through the air. At maximum ___________, the ___________ component of velocity equals ___________. The range formula \(R = v^2\sin(2\theta)/g\) only applies when launch and landing heights are ___________. When calculating maximum height, you must use the vertical ___________ of the launch velocity, not the total speed.
4. Short recall questions
Answer each question in 1–2 sentences using precise physics terms. 8 marks (2 each)
4.1 State the two conditions that must be satisfied before you can use the range formula \(R = v^2\sin(2\theta)/g\).
4.2 Explain why a student who writes \(v_x = v\) (using total speed as horizontal velocity) will always get the wrong answer.
4.3 What sign should acceleration \(g\) take in vertical equations when the positive direction is defined as upward? Justify your answer.
4.4 Two projectiles are launched at the same speed from level ground at angles of 25° and 65° respectively. Without calculating, state whether their ranges are the same or different, and give the reason.
5. Complete the formula table
Fill in the missing variable name, unit, or symbol for each row. 10 marks (1 each)
| # | Symbol | Variable name | SI unit | Found using formula… |
|---|---|---|---|---|
| 5.1 | \(v_x\) | m s−1 | \(v_x = v\cos\theta\) | |
| 5.2 | \(v_{y0}\) | Initial vertical velocity component | \(v_{y0} = v\sin\theta\) | |
| 5.3 | \(R\) | Horizontal range (level ground) | m | |
| 5.4 | Total flight time (level ground) | s | \(T = 2v_{y0}/g\) | |
| 5.5 | \(h_{\max}\) | m | \(h_{\max} = v_{y0}^2/(2g)\) |
Q1 — Formula–condition match
A–ii B–v C–iii D–iv E–i F–vi
Q2 — True / false with correction
2.1 True.
2.2 False. The range formula only applies when launch and landing heights are equal (level ground). From a cliff, use \(y = v_{y0}t + \tfrac{1}{2}gt^2\) to find flight time, then \(x = v_x t\) for range.
2.3 True.
2.4 False. 30° and 60° give the same range (complementary angles) but different maximum heights. \(h \propto \sin^2\theta\), so \(\sin^2 60° = 0.75 \neq \sin^2 30° = 0.25\). The 60° projectile reaches three times the maximum height of the 30° projectile.
2.5 True.
2.6 False. When upward is positive, gravity acts downward so \(a = -9.8\) m s−2. Using \(+9.8\) reverses the direction of acceleration and gives incorrect answers.
Q3 — Cloze paragraph
In order: gravity / constant / parabolic / height / vertical / zero / equal / component.
Q4.1 — Conditions for range formula
(1) Launch and landing heights must be equal (level ground). (2) Air resistance is negligible (constant downward acceleration only).
Q4.2 — Using total speed as \(v_x\)
The launch velocity is at an angle \(\theta\) to the horizontal. The horizontal component \(v_x = v\cos\theta\) is always less than the total speed \(v\) (unless \(\theta = 0\)). Using \(v\) instead of \(v\cos\theta\) overestimates the horizontal velocity, giving a range and horizontal displacement that are too large.
Q4.3 — Sign of \(g\)
When upward is positive, \(g = -9.8\) m s−2 (negative). Gravity acts downward, which is the negative direction in this convention. Entering \(+9.8\) would imply gravity acts upward, reversing all vertical motion results.
Q4.4 — Complementary angles
The ranges are the same. Since 25° + 65° = 90°, they are complementary angles. From \(R = v^2\sin(2\theta)/g\): \(\sin(2 \times 25°) = \sin 50°\) and \(\sin(2 \times 65°) = \sin 130° = \sin 50°\). The same \(\sin\) value gives the same range.
Q5 — Formula table
5.1 Horizontal velocity component. 5.2 m s−1. 5.3 \(R = v^2\sin(2\theta)/g\) (level ground only). 5.4 \(T\). 5.5 Maximum height above launch point.