Physics • Year 12 • Module 5 • Lesson 5

Practical Investigation: Validating Projectile Motion

Build fluency with the horizontal-launch projectile formulas — carrying units through every line and reporting answers to the correct number of significant figures.

Build · Formula Fluency

1. Formula recall card

The five relationships below are everything you need for a horizontal-launch projectile (launch angle = 0°). For each formula, write the SI units of every variable and a one-line note on when to use it. 5 marks

Name	Formula	Variables & SI units
Vertical drop under gravity (from rest)	s_y = ½ g t²	s_y = ____ (____), g = 9.8 ____ (____), t = ____ (____)
Time of flight (from launch height h)	t = √(2h / g)	t = ____ (____), h = ____ (____), g = ____ (____)
Horizontal range	R = v_x · t	R = ____ (____), v_x = ____ (____), t = ____ (____)
Range from launch height (combined)	R = v_x √(2h / g)	R, v_x, h in SI; g = 9.8 m s⁻²
Theoretical gradient of R vs √h	m_theory = v_x √(2 / g)	m has units of ____ ; v_x in m s⁻¹; g in m s⁻²

Stuck? Revisit lesson § Card 1 "Aim and Theoretical Background" — the formula panel and the line "If v_x and g are constant, then R ∝ √h".

2. Worked example — finding the range from launch height

A steel ball bearing rolls off a horizontal ramp at v_x = 1.40 m s⁻¹. The exit point is h = 0.450 m above the floor. Use g = 9.80 m s⁻². Find the horizontal range R. Carry units through every line and round to 3 significant figures.

Step 1 — Identify the unknown. Find R.

Step 2 — Choose the formula.

Horizontal launch ⇒ use R = v_x √(2h / g) (combines vertical time of flight with horizontal motion).

Step 3 — Substitute with units.

R = (1.40 m s⁻¹) × √((2 × 0.450 m) / (9.80 m s⁻²))

R = (1.40 m s⁻¹) × √(0.900 m / 9.80 m s⁻²)

R = (1.40 m s⁻¹) × √(0.09184 s²)

R = (1.40 m s⁻¹) × (0.3030 s)

Step 4 — Evaluate and check units.

R = 0.4243 m. Units: (m s⁻¹) × (s) = m. ✓

Step 5 — Round to 3 significant figures.

Inputs have 3 sig figs (1.40, 0.450, 9.80), so the answer also gets 3 sig figs.

R = 0.424 m

3. Eight single-formula calculations

Use g = 9.80 m s⁻² throughout. Show every substitution with units; quote the final answer to the correct number of significant figures (match the least precise input). 2 marks each — 16 marks

Foundation — clean numbers (Q3.1 – Q3.3)

Q3.1 A ball is dropped from rest from a height of h = 1.25 m. Using s_y = ½ g t², find the time to hit the floor. 2 marks

Q3.2 A ball bearing leaves a horizontal ramp at v_x = 2.00 m s⁻¹ with time of flight t = 0.500 s. Using R = v_x·t, find the horizontal range. 2 marks

Q3.3 Using t = √(2h / g), find the time of flight for a horizontal launch from h = 0.800 m. 2 marks

Standard — typical HSC practical numbers (Q3.4 – Q3.6)

Q3.4 A ball bearing rolls off a ramp at v_x = 1.20 m s⁻¹; the exit is h = 0.300 m above the floor. Use R = v_x √(2h / g) to find the horizontal range. 2 marks

Q3.5 In a trial the measured range is R = 0.55 m from launch height h = 0.30 m. Use R = v_x √(2h / g), rearranged for v_x, to find the launch speed. 2 marks

Q3.6 A student plots R against √h and finds an experimental gradient m_exp = 0.452 m^½. Use v_x = m_exp √(g / 2) to find the launch speed implied by the gradient. 2 marks

Extension — multi-step single formula manipulation (Q3.7 – Q3.8)

Q3.7 A horizontal-launch projectile with v_x = 1.50 m s⁻¹ lands R = 0.642 m from the foot of the launch table. Using R = v_x √(2h / g), find the launch height h. Report to 3 sig figs. 2 marks

Q3.8 A student records R = 0.71 m at h = 0.50 m and R = 0.45 m at h = 0.20 m (lesson sample data). Calculate v_x from each data point using R = v_x √(2h / g). Are the two values consistent to within ±5%? 2 marks

Stuck? Revisit lesson § Card 3 "Results Table" — sample data — and § Card 4 "Analysis" Step 4 for the rearranged formula.

Answers — Do not peek before attempting

Q1 — Formula recall card (sample completed entries)

s_y = ½ g t²: s_y = vertical drop (m), g = 9.8 m s⁻², t = time (s). Use when: the object starts from rest vertically (true for any horizontal launch, because v_y0 = 0).
t = √(2h / g): t = time of flight (s), h = launch height (m), g = 9.8 m s⁻². Use when: you know the launch height and need the time to hit the floor for a horizontal launch.
R = v_x·t: R = horizontal range (m), v_x = horizontal launch speed (m s⁻¹), t = time of flight (s). Use when: v_x is constant (no air resistance) and you already have t.
R = v_x √(2h / g): Combined form — single step from h and v_x to R.
m_theory = v_x √(2 / g): Units of m: (m s⁻¹) × √(s² / m) = (m s⁻¹) × (s / √m) = √m = m^½. Use when: comparing the gradient of a R vs √h graph to theory (lesson Card 4 Step 2).

Q3.1 — Time to fall 1.25 m from rest

s_y = ½ g t² ⇒ t = √(2 s_y / g) = √((2 × 1.25 m) / (9.80 m s⁻²)) = √(0.2551 s²) = 0.5051 s.

Inputs: 3 sig figs ⇒ t = 0.505 s. Units: √(m / (m s⁻²)) = √(s²) = s ✓.

Q3.2 — Range from v_x and t

R = v_x · t = (2.00 m s⁻¹) × (0.500 s) = 1.00 m.

Inputs: 3 sig figs ⇒ R = 1.00 m. Units: (m s⁻¹) × (s) = m ✓.

Q3.3 — Time of flight from h = 0.800 m

t = √(2h / g) = √((2 × 0.800 m) / (9.80 m s⁻²)) = √(0.1633 s²) = 0.4041 s.

t = 0.404 s (3 sig figs).

Q3.4 — Range from v_x = 1.20 m s⁻¹, h = 0.300 m

R = v_x √(2h / g) = (1.20 m s⁻¹) √((2 × 0.300 m) / (9.80 m s⁻²))

= (1.20 m s⁻¹) √(0.06122 s²) = (1.20 m s⁻¹) × (0.2474 s) = 0.2969 m.

R = 0.297 m (3 sig figs). Units check: (m s⁻¹)(s) = m ✓.

Q3.5 — Launch speed from R = 0.55 m, h = 0.30 m

Rearrange: v_x = R / √(2h / g) = R √(g / 2h).

v_x = (0.55 m) × √((9.80 m s⁻²) / (2 × 0.30 m)) = (0.55 m) × √(16.33 s⁻²) = (0.55 m) × (4.041 s⁻¹) = 2.223 m s⁻¹.

Inputs have 2 sig figs ⇒ v_x = 2.2 m s⁻¹. Units: m × s⁻¹ = m s⁻¹ ✓.

Q3.6 — Launch speed from experimental gradient

v_x = m_exp √(g / 2) = (0.452 m^½) × √((9.80 m s⁻²) / 2) = (0.452 m^½) × √(4.90 m s⁻²) = (0.452 m^½) × (2.214 m^½ s⁻¹) = 1.001 m s⁻¹.

v_x = 1.00 m s⁻¹ (3 sig figs). Units: (m^½) × (m^½ s⁻¹) = m s⁻¹ ✓.

Q3.7 — Launch height from R = 0.642 m, v_x = 1.50 m s⁻¹

R = v_x √(2h / g) ⇒ (R / v_x)² = 2h / g ⇒ h = g (R / v_x)² / 2.

R / v_x = (0.642 m) / (1.50 m s⁻¹) = 0.4280 s.

h = (9.80 m s⁻²) × (0.4280 s)² / 2 = (9.80 m s⁻²) × (0.1832 s²) / 2 = 0.8975 m.

h = 0.897 m (3 sig figs). Units: (m s⁻²) × (s²) = m ✓.

Q3.8 — Consistency check from two data points

Point A: R = 0.71 m, h = 0.50 m.

v_x,A = R √(g / 2h) = (0.71 m) × √((9.80 m s⁻²) / (2 × 0.50 m)) = (0.71 m) × √(9.80 s⁻²) = (0.71 m) × (3.130 s⁻¹) = 2.22 m s⁻¹.

Point B: R = 0.45 m, h = 0.20 m.

v_x,B = (0.45 m) × √((9.80 m s⁻²) / (2 × 0.20 m)) = (0.45 m) × √(24.5 s⁻²) = (0.45 m) × (4.950 s⁻¹) = 2.23 m s⁻¹.

Both round to v_x ≈ 2.2 m s⁻¹ (2 sig figs, matching inputs). % difference = |2.23 − 2.22| / 2.22 × 100% = 0.45%, well within ±5% ⇒ consistent.

Marking note: the consistency itself is the validation of the model — for a horizontal launch, the implied v_x from R = v_x√(2h/g) should be the same at every height because v_x is a controlled variable. Both data points give v_x ≈ 2.2 m s⁻¹, consistent with the lesson’s stated sample-data assumption of v_x ≈ 2.2 m s⁻¹.