Projectile Motion

This lesson covers AQA Spec 3.4.1.3 in detail. A projectile is any object moving freely under gravity with no driving force. The key principle is that horizontal and vertical motions are completely independent.

Spec mapping (AQA 7408 §3.4.1.3): This lesson develops the independence of horizontal and vertical motion under gravity, the trajectory of objects projected horizontally and at an angle, and the qualitative effects of air resistance on real projectile flight. The mathematical engine is the constant-acceleration SUVAT framework from §3.4.1.2 applied component-wise. Candidates are expected to be able to derive standard results (range, time of flight, maximum height) and to handle launches from height where the trajectory is asymmetric. (Refer to the official AQA specification document for exact wording.)

Synoptic links: Projectile motion is the cleanest A-Level example of vector decomposition into independent axes — a technique that recurs everywhere. (i) Newton's second law on inclined planes (Lesson 3, §3.4.1.4) uses the same axis-choice strategy: resolve gravity into "along the slope" and "perpendicular to slope" components. (ii) Momentum in two dimensions (Lesson 4, §3.4.1.6) decomposes the conservation equation into independent x- and y-component conservations, exactly as projectile motion decomposes acceleration. (iii) Electric and magnetic field problems in Year 13 treat charged-particle trajectories in deflection fields using identical machinery — uniform horizontal field, gravity-like deflection. The "independent components" idea is the conceptual core of mechanics, not merely a trick for cannonballs.

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1 The Independence of Horizontal and Vertical Motion

When air resistance is negligible:

• Horizontal motion: No force acts horizontally → zero acceleration → constant horizontal velocity.
  • x = vₓ t, where vₓ = u cos θ
• Vertical motion: Only gravity acts → constant acceleration g downward.
  • Apply SUVAT with a = −g (taking up as positive)

These two motions happen simultaneously and are linked only by the time variable t.

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2 Horizontal Launch (θ = 0°)

An object projected horizontally from a height h with speed u:

• Horizontal: x = ut
• Vertical: h = ½gt² (dropped from rest vertically)

Worked Example 1 — Ball Rolled Off a Table

A ball rolls off a table 1.25 m high with a horizontal speed of 3.0 m s⁻¹. Find (a) the time to reach the ground and (b) the horizontal distance from the base of the table.

Solution:

(a) Vertical: s = ½gt² → 1.25 = ½ × 9.81 × t² → t² = 2.50/9.81 = 0.2548 → t = 0.505 s

(b) Horizontal: x = ut = 3.0 × 0.505 = 1.51 m

(c) Speed at impact:

• vₓ = 3.0 m s⁻¹ (unchanged)
• vᵧ = gt = 9.81 × 0.505 = 4.95 m s⁻¹
• v = √(3.0² + 4.95²) = √(9.0 + 24.5) = √33.5 = 5.79 m s⁻¹
• Angle below horizontal: θ = tan⁻¹(4.95/3.0) = 58.8°

Exam Tip: For a horizontal launch, the vertical initial velocity is zero. Do not use the horizontal speed in the vertical SUVAT equations.

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3 Angled Launch — Full Treatment

For a projectile launched at speed u at angle θ above the horizontal from ground level:

• Horizontal component: uₓ = u cos θ
• Vertical component: uᵧ = u sin θ

Time of Flight

At the highest point, vᵧ = 0. Time to reach the top:

0 = uᵧ − gt_top → t_top = u sin θ / g

For a launch and landing at the same height (symmetrical trajectory):

T = 2u sin θ / g

Maximum Height

At the top: vᵧ² = uᵧ² − 2gH → 0 = u² sin² θ − 2gH

H = u² sin² θ / (2g)

Range

The horizontal distance for a complete flight:

R = uₓ × T = u cos θ × 2u sin θ / g

Using the identity 2 sin θ cos θ = sin 2θ:

R = u² sin 2θ / g

Key Result: Maximum range occurs when sin 2θ = 1, i.e., 2θ = 90° → θ = 45° (in the absence of air resistance).

Complementary Angles

Since sin 2θ = sin(180° − 2θ), two launch angles give the same range: θ and (90° − θ). For example, 30° and 60° give equal ranges.

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4 Worked Examples — Angled Launch

Worked Example 2

A javelin is thrown at 25 m s⁻¹ at 35° above the horizontal from ground level. Find (a) time of flight, (b) maximum height, (c) range. Take g = 9.81 m s⁻².

Solution:

uₓ = 25 cos 35° = 25 × 0.8192 = 20.48 m s⁻¹

uᵧ = 25 sin 35° = 25 × 0.5736 = 14.34 m s⁻¹

(a) T = 2uᵧ/g = 2 × 14.34 / 9.81 = 28.68/9.81 = 2.92 s

(b) H = uᵧ² / (2g) = 14.34² / (2 × 9.81) = 205.6 / 19.62 = 10.5 m

(c) R = uₓ × T = 20.48 × 2.92 = 59.8 m

Check with formula: R = u² sin 2θ / g = 625 × sin 70° / 9.81 = 625 × 0.9397 / 9.81 = 59.8 m ✓

Worked Example 3 — Finding the Velocity at Any Time

Using the javelin from Example 2, find the velocity (magnitude and direction) at t = 2.0 s.

Solution:

• vₓ = uₓ = 20.48 m s⁻¹ (constant)
• vᵧ = uᵧ − gt = 14.34 − 9.81 × 2.0 = 14.34 − 19.62 = −5.28 m s⁻¹ (downward)
• Speed: v = √(20.48² + 5.28²) = √(419.4 + 27.9) = √447.3 = 21.1 m s⁻¹
• Direction: θ = tan⁻¹(5.28/20.48) = 14.5° below the horizontal

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5 Launch From a Height

When a projectile is launched from height h above the landing level, the trajectory is no longer symmetrical.

Worked Example 4 — Launch From a Cliff

A stone is thrown at 18 m s⁻¹ at 25° above the horizontal from the top of a 45 m cliff. Find the time to reach the sea below and the horizontal distance from the base of the cliff.

Solution: Take upward as positive, origin at the launch point.

uₓ = 18 cos 25° = 18 × 0.9063 = 16.31 m s⁻¹

uᵧ = 18 sin 25° = 18 × 0.4226 = 7.61 m s⁻¹

The stone lands when the vertical displacement s = −45 m (below the launch point).

s = uᵧt + ½(−g)t²

−45 = 7.61t − 4.905t²

4.905t² − 7.61t − 45 = 0

Using the quadratic formula: t = [7.61 ± √(57.9 + 882.9)] / 9.81 = [7.61 ± √940.8] / 9.81

t = [7.61 ± 30.67] / 9.81

Taking the positive root: t = 38.28 / 9.81 = 3.90 s

Horizontal distance: x = uₓ × t = 16.31 × 3.90 = 63.6 m

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6 The Effect of Air Resistance

In reality, air resistance acts on a projectile. Its effects include:

• The maximum height is reduced (drag opposes upward motion)
• The range is reduced (drag decelerates the projectile throughout)
• The trajectory is asymmetric — the descending path is steeper than the ascending path
• The optimal launch angle for maximum range is less than 45° (typically around 40–42° depending on the object)
• Terminal velocity may be reached during descent

Exam Tip: AQA may ask you to sketch or describe the trajectory with air resistance compared to without. Always note that the range and maximum height are both reduced, and the path is no longer a symmetrical parabola.

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7 Experimental Determination of g Using Projectile Motion

Method: Roll a ball off a bench of known height h. Measure the horizontal distance x from the base.

From h = ½gt²: t = √(2h/g)

From x = uₓt: uₓ = x/t

Plot x² against h: since x = uₓ√(2h/g), we get x² = 2uₓ²h/g.

The gradient = 2uₓ²/g, so g = 2uₓ²/gradient.

AQA Specification Reference: Section 3.4.1.3 (Projectile motion). Students should be able to analyse projectile motion for objects launched horizontally or at an angle, and understand the independence of horizontal and vertical components.

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8 Worked Example 5 — Clearing an Obstacle (Extended)

A ball is kicked from ground level at 22 m s⁻¹ at 40° above the horizontal. A wall 2.0 m tall stands 35 m away. Determine whether the ball clears the wall and, if so, by how much.

Solution:

uₓ = 22 cos 40° = 22 × 0.766 = 16.85 m s⁻¹

uᵧ = 22 sin 40° = 22 × 0.643 = 14.14 m s⁻¹

Time to reach the horizontal position x = 35 m:

t = x / uₓ = 35 / 16.85 = 2.077 s

Vertical position at this time:

y = uᵧt − ½gt² = 14.14 × 2.077 − ½ × 9.81 × 2.077² = 29.37 − 21.16 = 8.21 m

Since 8.21 m > 2.0 m, the ball clears the wall by 8.21 − 2.0 = 6.21 m.

Check using range formula: Maximum range R = u² sin 2θ / g = 22² × sin 80° / 9.81 = 484 × 0.985 / 9.81 = 48.6 m. The wall is at 35 m, which is comfortably within the maximum range; the result is physically reasonable.

Worked Example 6 — Finding the Launch Angle for a Required Range

A projectile is launched from ground level with speed 30 m s⁻¹ and must land exactly 50 m away. Find the two possible launch angles. Take g = 9.81 m s⁻².

Solution: Use R = u² sin 2θ / g.

50 = 30² × sin 2θ / 9.81 = 900 sin 2θ / 9.81

sin 2θ = 50 × 9.81 / 900 = 0.5450

2θ = sin⁻¹(0.5450) = 33.04° or 2θ = 180° − 33.04° = 146.96°

Therefore θ = 16.5° (a flat, fast trajectory) or θ = 73.5° (a high, lobbed trajectory).

Both angles deliver the same horizontal range because they are complementary (16.5° + 73.5° = 90°). The flatter trajectory has a shorter time of flight; the lobbed trajectory has a longer time of flight and higher peak. This is the principle behind why a cricket fielder can choose a flat throw or a high lob to cover the same ground distance.

Worked Example 7 — Projectile Landing on a Sloping Surface

A stone is launched from the foot of a slope at 20 m s⁻¹ at 60° above the horizontal. The slope rises at 30° above the horizontal. Find the distance along the slope at which the stone lands.

Solution: Choose axes along the slope (positive uphill) and perpendicular to the slope. The launch velocity components relative to these axes:

u_along = 20 cos(60° − 30°) = 20 cos 30° = 20 × 0.866 = 17.32 m s⁻¹

u_perp = 20 sin(60° − 30°) = 20 sin 30° = 20 × 0.50 = 10.00 m s⁻¹

Gravity components relative to slope axes:

g_along = −g sin 30° = −9.81 × 0.50 = −4.905 m s⁻² (deceleration along the slope)

g_perp = −g cos 30° = −9.81 × 0.866 = −8.496 m s⁻² (decelerates the perpendicular component, drawing the stone back toward the slope)

Time to land on the slope: the perpendicular displacement returns to zero.

s_perp = u_perp × t + ½ × g_perp × t² = 0

10.00 t − 4.248 t² = 0

t(10.00 − 4.248 t) = 0

Reject t = 0 (launch instant). Therefore t = 10.00/4.248 = 2.354 s.

Distance along the slope:

s_along = u_along × t + ½ × g_along × t² = 17.32 × 2.354 + ½ × (−4.905) × 2.354²

s_along = 40.77 − 13.59 = 27.18 m (along the slope).

This is an extension-difficulty problem because resolving along the slope (rather than horizontal/vertical) makes the geometry tractable: the perpendicular axis sees uniform acceleration with the perpendicular velocity reversing, just like in horizontal projectile motion. The technique generalises to any sloping surface and is the classical approach to ski-jump and ramp-trajectory problems.

Worked Example 8 — Time to Reach Maximum Height for a Given Range