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.

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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