Physics • Year 12 • Module 5 • Lesson 5

Practical Investigation: Validating Projectile Motion

Apply the horizontal-launch model by combining time-of-flight, range and component formulas with a labelled diagram for each scenario.

Apply · Multi-Step Problem

1. Worked multi-step example — landing-velocity vector

A ball bearing leaves a horizontal ramp on a bench at v_x = 1.40 m s⁻¹. The exit point is h = 0.800 m above the floor. Take g = 9.80 m s⁻²; ignore air resistance. Find:

the time of flight,
the horizontal range,
the magnitude and direction of the velocity at the moment of impact.

Step 0 — Draw the setup.

Step 1 — Time of flight (vertical motion, starts from rest vertically).

s_y = ½ g t² ⇒ 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).

Step 2 — Horizontal range.

v_x is constant (no air resistance), so R = v_x·t.

R = (1.40 m s⁻¹) × (0.4041 s) = 0.5658 m ⇒ R = 0.566 m.

Step 3 — Vertical landing speed.

v_y = g·t = (9.80 m s⁻²) × (0.4041 s) = 3.960 m s⁻¹ downward.

Step 4 — Resultant landing velocity (vector addition).

|v| = √(v_x² + v_y²) = √((1.40)² + (3.960)²) m s⁻¹ = √(1.960 + 15.68) m s⁻¹ = √(17.64) m s⁻¹ = 4.200 m s⁻¹.

θ = tan⁻¹(v_y / v_x) = tan⁻¹(3.960 / 1.40) = tan⁻¹(2.829) = 70.5° below horizontal.

v = 4.20 m s⁻¹ at 70.5° below the horizontal (3 sig figs throughout).

2. Four multi-step problems

Each problem requires 2–3 formulas from the lesson plus a labelled sketch. Draw your sketch in the sketch box, show every substitution with units, and quote final answers to 3 significant figures (or the precision of the inputs, whichever is fewer). Use g = 9.80 m s⁻².

Q2.1 — Cliff launch 6 marks

A ball is thrown horizontally at v_x = 12.0 m s⁻¹ from a cliff h = 24.0 m high (from lesson Activity 1, Q3).

(i) Sketch the trajectory. Label v_x, h and R. (ii) Find the time of flight. (iii) Find the horizontal range. (iv) Find the magnitude and direction of the velocity 1.00 s after launch.

Sketch box — draw labelled diagram here

Stuck? Revisit lesson § Card 1 (formula panel — t and R) and Activity 1, Q3.

Q2.2 — Decomposing an oblique launch 6 marks

A projectile is launched at u = 24.0 m s⁻¹ at 44° above the horizontal (from lesson Activity 1, Q1). For this single question only, treat the launch as oblique (not horizontal).

(i) Sketch the launch with u, the angle, u_x and u_y labelled. (ii) Calculate u_x and u_y. (iii) Calculate the time to reach the maximum height (where v_y = 0) using v_y = u_y − g·t. (iv) Calculate the maximum height above the launch point.

Sketch box — draw labelled launch triangle here

Stuck? Revisit lesson Activity 1, Q1 ("Find the horizontal and vertical components of the initial velocity") and Activity 2 (independence of horizontal and vertical motion).

Q2.3 — Resultant velocity from components 5 marks

At a point in flight a projectile has horizontal velocity component v_x = 16.0 m s⁻¹ and vertical (downward) component v_y = 12.0 m s⁻¹ (from lesson Activity 1, Q2).

(i) Sketch v_x, v_y and the resultant v as a vector triangle. (ii) Calculate the magnitude |v|. (iii) Calculate the direction of v relative to the horizontal. (iv) State whether the projectile is moving toward or away from the launch point and justify with one sentence.

Sketch box — draw labelled vector triangle here

Q2.4 — Confirming the validation gradient 7 marks

A student conducts the lesson's horizontal-launch practical. They release the ball bearing from the same fixed position on the ramp at five heights and obtain a graph of R against √h with experimental gradient m_exp = 0.94 m^½. The launch speed is independently measured by a photogate at the ramp exit to be v_x,photogate = 2.20 m s⁻¹.

(i) Sketch a labelled R vs √h graph showing the line of best fit and where the gradient is taken. (ii) Calculate the theoretical gradient m_theory = v_x √(2/g) using the photogate v_x. (iii) Calculate the percentage difference between m_exp and m_theory using the lesson formula. (iv) State whether the model is validated using the lesson's "<10% acceptable" criterion, and identify one source of error that could account for any discrepancy.

Sketch box — draw labelled R vs √h graph here

Stuck? Revisit lesson § Card 4 "Analysis and Validation" (Steps 2 and 3) and § Validation Criteria callout.

Answers — Do not peek before attempting

Q2.1 — Cliff launch (v_x = 12.0 m s⁻¹, h = 24.0 m)

(i) Sketch: Cliff of height h on the left, ball leaves the top horizontally with v_x arrow pointing right, parabolic curve sweeping down to the ground a horizontal distance R from the base, R arrow labelled along the ground. Mark v_x and v_y components at the landing point.

(ii) Time of flight:

t = √(2h / g) = √((2 × 24.0 m) / (9.80 m s⁻²)) = √(4.898 s²) = 2.213 s ⇒ t = 2.21 s.

(iii) Range:

R = v_x·t = (12.0 m s⁻¹) × (2.213 s) = 26.56 m ⇒ R = 26.6 m (3 sig figs).

(iv) Velocity at t = 1.00 s:

v_x = 12.0 m s⁻¹ (unchanged, no air resistance).

v_y = g·t = (9.80 m s⁻²) × (1.00 s) = 9.80 m s⁻¹ downward.

|v| = √((12.0)² + (9.80)²) m s⁻¹ = √(144 + 96.04) m s⁻¹ = √(240.04) m s⁻¹ = 15.49 m s⁻¹.

θ = tan⁻¹(9.80 / 12.0) = tan⁻¹(0.8167) = 39.2° below horizontal.

v = 15.5 m s⁻¹ at 39.2° below the horizontal.

Q2.2 — Oblique launch (u = 24.0 m s⁻¹ at 44°)

(i) Sketch: Right-angled triangle with hypotenuse u at angle 44° above horizontal, horizontal leg u_x, vertical leg u_y.

(ii) Components:

u_x = u cos(44°) = (24.0 m s⁻¹) × (0.7193) = 17.26 m s⁻¹ ⇒ u_x = 17.3 m s⁻¹.

u_y = u sin(44°) = (24.0 m s⁻¹) × (0.6947) = 16.67 m s⁻¹ ⇒ u_y = 16.7 m s⁻¹ upward.

(iii) Time to max height (where v_y = 0):

0 = u_y − g·t_top ⇒ t_top = u_y / g = (16.67 m s⁻¹) / (9.80 m s⁻²) = 1.701 s ⇒ t_top = 1.70 s.

(iv) Maximum height:

h_max = u_y·t_top − ½ g·t_top² = (16.67)(1.701) − ½(9.80)(1.701)² m

= 28.36 m − ½(9.80)(2.893) m = 28.36 m − 14.18 m = 14.18 m.

(Cross-check using v² = u² − 2g·h: h = u_y² / 2g = (16.67)² / (2 × 9.80) = 277.9 / 19.60 = 14.18 m ✓.)

h_max = 14.2 m above launch point (3 sig figs).

Q2.3 — Resultant velocity (v_x = 16.0 m s⁻¹, v_y = 12.0 m s⁻¹ down)

(i) Sketch: Horizontal arrow v_x = 16.0 m s⁻¹ to the right; vertical arrow v_y = 12.0 m s⁻¹ straight down from the tip; resultant v drawn from the tail of v_x to the tip of v_y.

(ii) Magnitude:

|v| = √((16.0)² + (12.0)²) m s⁻¹ = √(256 + 144) m s⁻¹ = √(400) m s⁻¹ = 20.0 m s⁻¹.

|v| = 20.0 m s⁻¹.

(iii) Direction:

θ = tan⁻¹(v_y / v_x) = tan⁻¹(12.0 / 16.0) = tan⁻¹(0.7500) = 36.9° below horizontal.

v = 20.0 m s⁻¹ at 36.9° below the horizontal.

(iv) Direction of motion: The projectile is moving away from the launch point. The horizontal component remains positive (forward) and the vertical component is downward, which corresponds to the descending half of the trajectory — the projectile cannot return horizontally because v_x is constant in the no-air-resistance model.

Q2.4 — Validation gradient (m_exp = 0.94 m^½, v_x,photogate = 2.20 m s⁻¹)

(i) Sketch: Axes labelled √h (m^½) on the x-axis, R (m) on the y-axis; line of best fit through origin and the data points; rise/run triangle drawn on the line to indicate where the gradient is calculated, with ΔR and Δ√h labelled.

(ii) Theoretical gradient:

m_theory = v_x √(2 / g) = (2.20 m s⁻¹) × √(2 / 9.80 s² m⁻¹) = (2.20 m s⁻¹) × √(0.2041 s² m⁻¹)

= (2.20 m s⁻¹) × (0.4518 s m^−½) = 0.9940 m^½.

Units check: (m s⁻¹) × (s m^−½) = m^½ ✓.

m_theory = 0.994 m^½ (3 sig figs).

(iii) Percentage difference:

% diff = |m_exp − m_theory| / m_theory × 100% = |0.94 − 0.994| / 0.994 × 100% = 0.054 / 0.994 × 100% = 5.43%.

% difference ≈ 5.4%.

(iv) Validation judgement: The percentage difference (5.4%) is well under the lesson's 10% acceptance threshold, so the model R = v_x√(2h/g) is validated by these data. A plausible source for the small remaining discrepancy is rolling friction on the ramp (slightly reducing the actual exit speed below the photogate reading averaged over the exit window), or a tiny non-zero ramp tilt at the exit (giving a small downward v_y at launch that shortens R compared with a pure horizontal launch). Both effects would push m_exp below m_theory, which is exactly the direction observed here.