26  Projectile Motion

26.1 Syllabus inquiry question

  • How can models be used to predict motion in two dimensions?
Feynman Insight

From The Feynman Lectures on Physics, Vol I, Chapter 8:

The genius of Galileo’s insight was realizing that the horizontal and vertical motions are independent. One does not affect the other. This simple idea unlocks all of projectile motion.

26.2 Learning Objectives

  • Analyse projectile motion by resolving into horizontal and vertical components.
  • Apply equations for time of flight, maximum height, and range.
  • Solve problems involving non-level launches.
  • Conduct practical investigations to validate projectile relationships.

26.3 Content

26.3.1 Model and Assumptions

A projectile is any object moving through the air under the influence of gravity alone.

Key Assumptions
  • Constant vertical acceleration: \(g = 9.8\ \text{m/s}^2\) downward
  • Zero air resistance: no drag forces
  • Independence of motion: horizontal and vertical motions do not affect each other

26.3.2 Interactive: Projectile Trajectory

Explore how launch angle and speed affect projectile motion:

26.3.3 Component Analysis

Split the initial velocity into horizontal and vertical components:

\[v_x = v_0 \cos(\theta) \quad \text{(constant)}\] \[v_{y0} = v_0 \sin(\theta) \quad \text{(changes with time)}\]

Horizontal motion (constant velocity): \[x = v_x t\]

Vertical motion (constant acceleration): \[v_y = v_{y0} - gt\] \[y = v_{y0}t - \frac{1}{2}gt^2\]

26.3.4 Time of Flight (Level Launch)

For a projectile launched and landing at the same height:

\[T = \frac{2v_0 \sin(\theta)}{g}\]

Understanding Time of Flight

Time of flight depends only on the vertical component of velocity. The horizontal velocity has no effect on how long the projectile stays in the air.

26.3.5 Maximum Height

The projectile reaches maximum height when \(v_y = 0\):

\[H = \frac{v_0^2 \sin^2(\theta)}{2g}\]

26.3.6 Interactive: Maximum Height Analysis

Key observation: At maximum height, the vertical velocity is zero but horizontal velocity remains constant.

26.3.7 Horizontal Range (Level Launch)

For level launch and landing:

\[R = \frac{v_0^2 \sin(2\theta)}{g}\]

Maximum Range

Maximum range occurs at \(\theta = 45°\). Complementary angles (e.g., 30° and 60°) give the same range.

26.3.8 Interactive: Range Comparison

Compare trajectories at different launch angles:

26.3.9 Non-Level Launch

When launch and landing heights differ, use the general vertical equation:

\[y = v_{y0}t - \frac{1}{2}gt^2\]

Set \(y\) to the final height (relative to launch) and solve for time.

26.4 Worked Examples

26.4.1 Example 1: Level launch calculations

A projectile is launched at \(25\ \text{m/s}\) at \(40°\) above horizontal. Find time of flight and range.

Solution:

  1. Time of flight: \(T = \frac{2v_0 \sin(\theta)}{g} = \frac{2 \times 25 \times \sin(40°)}{9.8}\)

  2. \(T = \frac{32.14}{9.8} = 3.28\ \text{s}\)

  3. Range: \(R = v_x \times T = 25\cos(40°) \times 3.28 = 19.15 \times 3.28 = 62.8\ \text{m}\)

26.4.2 Example 2: Maximum height

A ball is kicked at \(20\ \text{m/s}\) at \(30°\). Find the maximum height.

Solution:

  1. Use \(H = \frac{v_0^2 \sin^2(\theta)}{2g}\)

  2. \(H = \frac{20^2 \times \sin^2(30°)}{2 \times 9.8} = \frac{400 \times 0.25}{19.6}\)

  3. \(H = 5.1\ \text{m}\)

26.4.3 Example 3: Horizontal launch from height

A ball is thrown horizontally at \(8.0\ \text{m/s}\) from a height of \(20\ \text{m}\). Find time to impact and horizontal distance.

Solution:

  1. Vertical motion: \(y = -\frac{1}{2}gt^2\) (since \(v_{y0} = 0\))

  2. \(-20 = -\frac{1}{2} \times 9.8 \times t^2\), so \(t = 2.02\ \text{s}\)

  3. Horizontal distance: \(x = v_x t = 8.0 \times 2.02 = 16.2\ \text{m}\)

26.4.4 Example 4: Final velocity

Find the final velocity (magnitude and direction) for Example 3.

Solution:

  1. \(v_x = 8.0\ \text{m/s}\) (unchanged)

  2. \(v_y = gt = 9.8 \times 2.02 = 19.8\ \text{m/s}\) (downward)

  3. \(v = \sqrt{v_x^2 + v_y^2} = \sqrt{64 + 392} = 21.4\ \text{m/s}\)

  4. \(\theta = \tan^{-1}(19.8/8.0) = 68°\) below horizontal

26.5 Common Misconceptions

Common Misconceptions
  • Misconception: Horizontal and vertical motions affect each other. Correction: They are completely independent. Gravity only affects vertical motion.

  • Misconception: Time of flight depends on horizontal speed. Correction: Time of flight depends only on vertical component and height change.

  • Misconception: Heavier objects fall faster. Correction: Without air resistance, all objects accelerate at \(g\) regardless of mass.

  • Misconception: Velocity is zero at maximum height. Correction: Only vertical velocity is zero; horizontal velocity remains constant.

  • Misconception: The range formula works for all launches. Correction: \(R = v_0^2 \sin(2\theta)/g\) only applies when launch and landing heights are equal.

26.6 Practice Questions

26.6.1 Easy (2 marks)

A projectile is launched at \(15\ \text{m/s}\) at \(30°\). Calculate the horizontal and vertical components of initial velocity.

  • Use \(v_x = v_0 \cos(\theta)\) and \(v_y = v_0 \sin(\theta)\) (1)
  • Correct values: \(v_x = 13.0\ \text{m/s}\), \(v_y = 7.5\ \text{m/s}\) (1)

Answer: \(v_x = 13\ \text{m/s}\), \(v_y = 7.5\ \text{m/s}\)

26.6.2 Medium (4 marks)

A ball is kicked at \(25\ \text{m/s}\) at \(40°\). Calculate the time of flight and maximum height.

  • Time of flight formula: \(T = 2v_0 \sin(\theta)/g\) (1)
  • Correct time: \(T = 3.28\ \text{s}\) (1)
  • Maximum height formula: \(H = v_0^2 \sin^2(\theta)/(2g)\) (1)
  • Correct height: \(H = 13.2\ \text{m}\) (1)

Answer: \(T = 3.3\ \text{s}\), \(H = 13\ \text{m}\)

26.6.3 Hard (5 marks)

A stone is thrown at \(18\ \text{m/s}\) at \(35°\) from a cliff \(45\ \text{m}\) high. Find the time to hit the ground and the horizontal distance from the base of the cliff.

  • Identify \(v_{y0} = 18 \sin(35°) = 10.3\ \text{m/s}\) (1)
  • Set up: \(-45 = 10.3t - 4.9t^2\) (1)
  • Solve quadratic: \(t = 4.24\ \text{s}\) (positive root) (1)
  • Calculate \(v_x = 18 \cos(35°) = 14.7\ \text{m/s}\) (1)
  • Range: \(x = 14.7 \times 4.24 = 62.3\ \text{m}\) (1)

Answer: \(t = 4.2\ \text{s}\), \(x = 62\ \text{m}\)

26.7 Multiple Choice Questions

Test your understanding with these interactive questions:

26.8 Quick Quiz: Projectile Motion

Test your understanding of projectile motion with this timed quiz:

26.9 Extended Response Practice

26.10 Summary

Key Takeaways
  • Horizontal motion: constant velocity (\(x = v_x t\))
  • Vertical motion: constant acceleration (\(v_y = v_{y0} - gt\))
  • Time of flight (level): \(T = 2v_0 \sin(\theta)/g\)
  • Maximum height: \(H = v_0^2 \sin^2(\theta)/(2g)\)
  • Range (level): \(R = v_0^2 \sin(2\theta)/g\)
  • Maximum range occurs at \(\theta = 45°\)

26.11 Self-Assessment

Check your understanding:

After studying this section, you should be able to: