Projectile Motion

Introduction to Projectile Motion

• A projectile is any object that is launched, thrown, or otherwise projected into the air and subject only to the acceleration of gravity and air resistance.
• Projectile motion is a two-dimensional motion that can be analyzed by considering the horizontal and vertical components of the motion separately.
• We often make the simplifying assumption that air resistance is negligible.

KEY TAKEAWAY: Projectile motion is analyzing motion in two dimensions, separating it into horizontal and vertical components for easier calculation.

Analyzing Projectile Motion: Ignoring Air Resistance

Key Assumptions and Principles

• Neglecting Air Resistance: In ideal projectile motion, we assume that air resistance is negligible. This simplifies the analysis significantly.
• Gravity is the Only Force: The only force acting on the projectile is the force due to gravity, $F_g$.
• Constant Vertical Acceleration: This results in a constant downward acceleration of $g \approx 9.8 \, \text{m/s}^2$ (acceleration due to gravity).
• Independent Motion: The horizontal and vertical components of motion are independent of each other.
• Constant Horizontal Velocity: Since there is no horizontal force (ignoring air resistance), the horizontal velocity remains constant throughout the projectile’s flight.

Resolving Initial Velocity into Components

• The initial velocity, $u$, of a projectile launched at an angle $\theta$ to the horizontal can be resolved into horizontal and vertical components:
  • Horizontal component: $u_x = u \cos{\theta}$
  • Vertical component: $u_y = u \sin{\theta}$

Horizontal Motion

• Constant Velocity: The horizontal velocity ($v_x$) remains constant throughout the motion: $v_x = u_x = u \cos{\theta}$.
• Horizontal Displacement: The horizontal displacement ($s_x$) at time $t$ is given by:
  $$s_x = u_x t = (u \cos{\theta}) t$$

Vertical Motion

• Constant Acceleration: The vertical motion is subject to constant acceleration due to gravity ($a_y = -g$).
• Equations of Motion: Use the constant acceleration equations (SUVAT) to analyze the vertical motion:
  • $v_y = u_y + a_y t = u \sin{\theta} - gt$
  • $s_y = u_y t + \frac{1}{2} a_y t^2 = (u \sin{\theta}) t - \frac{1}{2} g t^2$
  • $v_y^2 = u_y^2 + 2 a_y s_y = (u \sin{\theta})^2 - 2 g s_y$

Key Parameters

• Time of Flight (T): The total time the projectile spends in the air. It can be found by setting $s_y = 0$ and solving for $t$:
  $$T = \frac{2 u \sin{\theta}}{g}$$
• Maximum Height (H): The maximum vertical displacement reached by the projectile. It occurs when $v_y = 0$:
  $$H = \frac{(u \sin{\theta})^2}{2g}$$
• Range (R): The total horizontal distance traveled by the projectile. It can be found by substituting the time of flight ($T$) into the horizontal displacement equation:
  $$R = \frac{u^2 \sin{2\theta}}{g}$$

Trajectory

• The trajectory of a projectile (ignoring air resistance) is a parabola.

EXAM TIP: Remember to always resolve the initial velocity into horizontal and vertical components. Correctly applying the SUVAT equations is crucial for solving projectile motion problems.

Analyzing Projectile Motion: With Air Resistance

Qualitative Description of Air Resistance

• Air Resistance: Air resistance is a force that opposes the motion of an object through the air. It depends on factors such as the object’s speed, shape, and the density of the air.
• Direction: Air resistance acts in the opposite direction to the velocity of the projectile.
• Effect on Horizontal Motion: Air resistance reduces the horizontal velocity of the projectile over time. The horizontal acceleration is no longer zero.
• Effect on Vertical Motion: Air resistance reduces the vertical velocity of the projectile, both on the way up and on the way down. The vertical acceleration is no longer just $-g$.

Effects of Air Resistance on Projectile Trajectory

• Reduced Range: Air resistance reduces the range of the projectile compared to the ideal case.
• Reduced Maximum Height: Air resistance reduces the maximum height reached by the projectile.
• Shorter Time of Flight: Air resistance reduces the time of flight of the projectile.
• Asymmetrical Trajectory: The trajectory is no longer a perfect parabola. The descending part of the trajectory is steeper than the ascending part.
• Terminal Velocity: As the projectile falls, the force of air resistance increases with speed. Eventually, the air resistance force equals the weight of the projectile, and the projectile reaches terminal velocity.

Qualitative Analysis

• Analyzing projectile motion with air resistance is complex and usually requires numerical methods or computer simulations.
• In VCE Physics, you are expected to provide a qualitative description of the effects of air resistance.

COMMON MISTAKE: Forgetting that air resistance acts in both the horizontal and vertical directions, causing deceleration in both.

Linking Horizontal and Vertical Motion

• Time as a Link: Time ($t$) is the common variable that links the horizontal and vertical motion of a projectile.
• Solving Problems:
  1. Resolve the initial velocity into horizontal and vertical components.
  2. Analyze the vertical motion to find the time of flight or the time to reach maximum height.
  3. Use this time to calculate the horizontal displacement (range).

STUDY HINT: Practice solving a variety of projectile motion problems, both with and without air resistance, to reinforce your understanding.

Formulas

• Constant acceleration equations (SUVAT):
  • $v = u + at$
  • $v^2 = u^2 + 2as$
  • $s = \frac{1}{2} (u + v) t$
  • $s = ut + \frac{1}{2} a t^2$
  • $s = vt - \frac{1}{2} a t^2$
• Average velocity:
  • $v = \frac{\Delta s}{\Delta t}$
• Horizontal motion:
  • $s_x = u_x t = (u \cos{\theta}) t$
• Vertical motion:
  • $v_y = u_y + a_y t = u \sin{\theta} - gt$
  • $s_y = u_y t + \frac{1}{2} a_y t^2 = (u \sin{\theta}) t - \frac{1}{2} g t^2$
  • $v_y^2 = u_y^2 + 2 a_y s_y = (u \sin{\theta})^2 - 2 g s_y$
• Time of Flight:
  $$T = \frac{2 u \sin{\theta}}{g}$$
• Maximum Height:
  $$H = \frac{(u \sin{\theta})^2}{2g}$$
• Range:
  $$R = \frac{u^2 \sin{2\theta}}{g}$$

REMEMBER: The range equation is only valid when the launch and landing heights are the same.

VCAA Examination Focus

• VCAA exams often include questions that require you to:
  • Apply the constant acceleration equations to solve projectile motion problems.
  • Explain the effects of air resistance on projectile motion.
  • Calculate the range, maximum height, and time of flight of a projectile (without air resistance).
  • Sketch the trajectory of a projectile with and without air resistance.
  • Analyze the horizontal and vertical components of projectile motion independently.

VCAA FOCUS: Expect questions that require you to explain why the range is affected by air resistance, not just state that it is reduced. Be prepared to discuss the changes in both horizontal and vertical velocities.

Examples

• A ball is thrown with an initial velocity of \$20 \, \text{m/s}$ at an angle of $30^\circ$ to the horizontal. Calculate the range, maximum height, and time of flight, ignoring air resistance.
• Describe how the range, maximum height, and time of flight would be affected if air resistance was taken into account.

APPLICATION: Understanding projectile motion is crucial in many real-world applications, such as sports (e.g., baseball, golf), military applications (e.g., artillery), and engineering (e.g., designing trajectories for rockets).