Physics • Year 12 • Module 5 • Lesson 3

Problem Solving: Projectiles

Lock in the six projectile formulas, the GIVEN→FIND→METHOD→ANSWER protocol, and the conditions under which the range equation applies before tackling harder problems.

Build · Vocab & Recall

1. Term–definition match

The definitions below are shuffled. In the right-hand column write the matching term from this list: range, time of flight, maximum height, horizontal component, vertical component, symmetrical flight, problem-solving protocol, trajectory, launch angle, projectile. 10 marks (1 each)

#DefinitionMatching term
1.1The horizontal distance from the launch point to the landing point.
1.2The total elapsed time from launch to landing.
1.3The peak vertical displacement above the launch point during a projectile’s flight.
1.4The part of the initial velocity that acts in the horizontal direction only; it remains constant throughout the flight (ignoring air resistance).
1.5The part of the initial velocity that acts in the vertical direction; it changes due to gravity.
1.6The special case of projectile motion in which the launch height equals the landing height, so the upward and downward portions of the path are mirror images.
1.7The structured approach to solving a physics problem: list knowns, state what to find, identify the equation, then substitute and solve.
1.8The curved path followed by a projectile under the influence of gravity.
1.9The angle between the initial velocity vector and the horizontal, measured in degrees.
1.10Any object launched into the air that moves under the influence of gravity only (neglecting air resistance).
Stuck? Revisit the Key Terms panel in the lesson’s Questions phase.

2. Fill-in-the-blank paragraph

Use the word bank to complete the passage. Each word or phrase is used once. 9 marks (1 per blank)

Word bank:

constant  ·  downward  ·  gravity  ·  horizontal  ·  level ground  ·  maximum height  ·  quadratic  ·  vertical  ·  zero

A projectile moves under the influence of ___________ only. The horizontal velocity is ___________ throughout the flight because no horizontal force acts. The ___________ velocity, however, changes at a rate of 9.8 m/s every second in the ___________ direction. Time of flight is determined entirely by ___________ motion. At the peak of the trajectory the vertical velocity is ___________, and this moment defines the ___________. The range equation $R = v^2\sin(2\theta)/g$ is valid only for ___________ scenarios. When launch and landing heights differ, the vertical equation of motion becomes a ___________ in time.

Stuck? Revisit Cards 1–3 of the lesson and the formula panel.

3. True or false — with correction

Circle T or F. If false, write the correct statement on the line below. 12 marks (1 T/F + 1 correction each)

3.1 The range equation $R = v^2\sin(2\theta)/g$ applies to any projectile problem regardless of launch and landing heights.    T  /  F

3.2 At the top of a projectile’s path, the acceleration due to gravity is momentarily zero.    T  /  F

3.3 Time of flight depends on the horizontal velocity of a projectile.    T  /  F

3.4 For a projectile launched on level ground, the total flight time equals twice the time to reach maximum height.    T  /  F

3.5 A heavier projectile will travel less far than a lighter projectile launched under identical conditions (same speed, angle, height), in the absence of air resistance.    T  /  F

3.6 When a projectile is launched from a height above the landing point, the vertical displacement is negative (taking upward as positive).    T  /  F

Stuck? Revisit the Common Misconceptions panel and the formula panel in the lesson.

4. Formula recall — purpose and units

For each formula below, state: (a) what physical quantity it calculates, and (b) the SI unit of the result. 8 marks (2 each)

4.1   $s_x = v\cos\theta \cdot t$

(a) What does it calculate?

(b) SI unit of the result:

4.2   $t_{\text{up}} = v\sin\theta / g$

(a) What does it calculate?

(b) SI unit of the result:

4.3   $h_{\text{max}} = (v\sin\theta)^2 / (2g)$

(a) What does it calculate?

(b) SI unit of the result:

4.4   $v_y = v\sin\theta - gt$

(a) What does it calculate?

(b) SI unit of the result:

Stuck? Revisit the formula panel (Key Formulas — Projectile Motion) in the lesson.

5. Build a concept map

Draw labelled arrows between the six terms below to show how they connect. Each arrow must carry a linking phrase (e.g. “is calculated using”, “determines”, “is independent of”). Aim for at least 6 labelled arrows. 6 marks (1 per valid arrow)

Supplied terms: launch speed · launch angle · horizontal component · vertical component · time of flight · range.

launch speed
launch angle
horizontal component
vertical component
time of flight
range
Hints: launch speed + launch angle → determine → horizontal component; vertical component → determines → time of flight; time of flight → used with horizontal component → to calculate → range.
Answers — Do not peek before attempting

Q1 — Term–definition match

1.1 range • 1.2 time of flight • 1.3 maximum height • 1.4 horizontal component • 1.5 vertical component • 1.6 symmetrical flight • 1.7 problem-solving protocol • 1.8 trajectory • 1.9 launch angle • 1.10 projectile.

Q2 — Cloze paragraph

In order: gravity / constant / vertical / downward / vertical / zero / maximum height / level ground / quadratic.

Q3 — True / false with correction

3.1 False. The range equation $R = v^2\sin(2\theta)/g$ applies only when launch and landing heights are equal (level ground). For uneven ground, use the equations of motion directly.

3.2 False. Gravity acts continuously at $g = 9.8\ \text{m/s}^2$ downward throughout the entire flight — including at the peak. Only the vertical velocity is momentarily zero at the top.

3.3 False. Time of flight is determined entirely by vertical motion (the initial vertical velocity and the height difference). The horizontal velocity affects range but not flight time.

3.4 True. In symmetrical flight the path is a mirror image about the peak, so $t_{\text{total}} = 2t_{\text{up}}$.

3.5 False. Projectile motion (neglecting air resistance) is independent of mass. Both objects follow the same trajectory under the same launch conditions — they land at the same point at the same time.

3.6 True. Taking upward as positive, a landing point below the launch point has a negative vertical displacement, so $s_y < 0$.

Q4 — Formula recall

4.1 (a) Horizontal displacement of the projectile during time $t$. (b) metres (m).

4.2 (a) Time taken for the projectile to reach maximum height (where $v_y = 0$). (b) seconds (s).

4.3 (a) Maximum vertical height above the launch point. (b) metres (m).

4.4 (a) Vertical component of velocity at time $t$ after launch. (b) metres per second (m/s).

Q5 — Sample concept map

Correct maps should include arrows such as:

  • launch speed + launch angletogether determinehorizontal component ($v_x = v\cos\theta$)
  • launch speed + launch angletogether determinevertical component ($v_y = v\sin\theta$)
  • vertical componentdeterminestime of flight
  • time of flightused with horizontal component to calculaterange
  • horizontal componentis independent oftime of flight
  • launch angleaffects bothhorizontal component and vertical component

Award 1 mark per valid labelled arrow (minimum 6).