Projectile Motion

Module: Advanced Mechanics

Overview

Projectile motion describes the motion of an object launched into the air that moves under the influence of gravity alone (ignoring air resistance). This is a key topic in Module 5 that builds on Year 11 kinematics.

Key Syllabus Points:

  • Analyse the motion of projectiles by resolving motion into horizontal and vertical components
  • Apply kinematic equations to projectile motion problems
  • Investigate projectile motion experimentally

Key Concepts

Independence of Horizontal and Vertical Motion

Fundamental Principle

The horizontal and vertical components of projectile motion are completely independent of each other. Gravity only affects the vertical component.

Component Acceleration Velocity
Horizontal \(a_x = 0\) \(v_x = u_x = u\cos\theta\) (constant)
Vertical \(a_y = -g = -9.8\) m/s² \(v_y = u_y + a_y t\) (changes)

Kinematic Equations

For horizontal motion (constant velocity): \[x = u_x t = (u\cos\theta)t\]

For vertical motion (constant acceleration): \[v_y = u_y + gt\] \[y = u_y t + \frac{1}{2}gt^2\] \[v_y^2 = u_y^2 + 2gy\]

Key Formulas Summary

Quantity Formula
Initial horizontal velocity \(u_x = u\cos\theta\)
Initial vertical velocity \(u_y = u\sin\theta\)
Time of flight \(T = \frac{2u\sin\theta}{g}\)
Maximum height \(H = \frac{u^2\sin^2\theta}{2g}\)
Range \(R = \frac{u^2\sin(2\theta)}{g}\)
Maximum range At \(\theta = 45°\)

Worked Examples

Example 1: Ball thrown horizontally

A ball is thrown horizontally at 15 m/s from a cliff 80 m high. Find: (a) Time to hit the ground (b) Horizontal distance travelled (c) Velocity on impact

Solution:

  1. Using \(y = u_y t + \frac{1}{2}gt^2\) with \(u_y = 0\): \[-80 = 0 + \frac{1}{2}(-9.8)t^2\] \[t = 4.04 \text{ s}\]

  2. Horizontal distance: \[x = u_x t = 15 \times 4.04 = 60.6 \text{ m}\]

  3. Final velocity components:

  • \(v_x = 15\) m/s (unchanged)
  • \(v_y = u_y + gt = 0 + (-9.8)(4.04) = -39.6\) m/s

Speed: \(v = \sqrt{v_x^2 + v_y^2} = \sqrt{15^2 + 39.6^2} = 42.3\) m/s

Angle: \(\theta = \tan^{-1}\left(\frac{39.6}{15}\right) = 69.3°\) below horizontal

Example 2: Projectile launched at an angle

A projectile is launched at 30 m/s at 40° above horizontal. Calculate: (a) Maximum height (b) Time of flight (c) Range

Solution:

  1. Maximum height: \[H = \frac{u^2\sin^2\theta}{2g} = \frac{30^2\sin^2(40°)}{2 \times 9.8} = 18.9 \text{ m}\]

  2. Time of flight: \[T = \frac{2u\sin\theta}{g} = \frac{2 \times 30 \times \sin(40°)}{9.8} = 3.94 \text{ s}\]

  3. Range: \[R = \frac{u^2\sin(2\theta)}{g} = \frac{30^2\sin(80°)}{9.8} = 90.4 \text{ m}\]

Common Misconceptions

Avoid These Mistakes
  1. Forgetting sign conventions - Choose a consistent direction as positive (usually up) and stick with it
  2. Using \(g = 10\) m/s² - In HSC, always use \(g = 9.8\) m/s² unless told otherwise
  3. Confusing components - Keep horizontal and vertical calculations separate
  4. Missing the projectile at different heights - Formulas for \(T\), \(H\), \(R\) assume launch and landing at same height
  5. Air resistance - In HSC Physics, assume negligible unless specifically stated

HSC Exam Analysis

Question Types

  1. Calculation questions (4-6 marks): Calculate time, range, height, or velocity
  2. Graphing questions (3-4 marks): Sketch trajectory or velocity-time graphs
  3. Analysis questions (5-7 marks): Compare trajectories, explain independence principle

Recent HSC Questions

  • 2024 Q21: Projectile calculation with non-level landing
  • 2023 Q24: Projectile analysis comparing two trajectories
  • 2022 Q23: Projectile launched from height

Practice Problems

  1. A ball is kicked at 25 m/s at 60° above horizontal. Calculate the maximum height and range.

  2. A projectile lands 100 m away after 4.0 s in the air. Calculate its initial speed and angle.

  3. Compare the trajectories of two projectiles launched at 30° and 60° with the same speed.

  4. A ball is thrown horizontally at 12 m/s from a 45 m cliff. How far from the base does it land?