# Projectile Motion ## Vertical & Horizontal Motion of a Projectile - Projectile motion describes the path of any object thrown or launched through the air - The trajectory can be split into two independent components: - **Vertical component**: subject to constant gravitational acceleration - **Horizontal component**: moves at constant velocity (assuming no air resistance) - These components must be analysed separately then combined to understand the full motion

The horizontal and vertical motions are completely independent of each other – what happens vertically doesn't affect horizontal motion and vice versa

### Key Parameters - **Time of flight**: the total duration the projectile remains airborne - **Maximum height**: the peak vertical position where vertical velocity becomes zero momentarily - **Range**: the total horizontal displacement from launch to landing ## Vertical Motion Analysis - Initial vertical velocity component: $u_y = u \sin \theta$ (where $\theta$ is the launch angle above horizontal) - Vertical acceleration: $a = -g = -9.81 \text{ m s}^{-2}$ (always downward) - At maximum height, vertical velocity $v = 0$ ### Vertical Velocity Changes Over Time The vertical velocity component changes linearly due to constant gravitational acceleration:

- Velocity decreases by $9.81 \text{ m s}^{-1}$ every second - At the peak, vertical velocity momentarily equals zero - After the peak, velocity becomes increasingly negative (downward) ### Time of Flight Calculation - For a projectile launched and landing at the same height, use: $$v = u + at$$ - At maximum height: $0 = u \sin \theta - gt_{\text{up}}$ - Time to reach maximum height: $t_{\text{up}} = \frac{u \sin \theta}{g}$ - Total time of flight: $t_{\text{total}} = \frac{2u \sin \theta}{g}$

Time of flight is twice the time taken to reach maximum height (for symmetric trajectories)

### Maximum Height Calculation - Use the equation: $v^2 = u^2 + 2as$ - At peak: $0 = (u \sin \theta)^2 - 2gH$ - Rearrange to find maximum height: $$H = \frac{(u \sin \theta)^2}{2g}$$ ## Horizontal Motion Analysis - Initial horizontal velocity component: $u_x = u \cos \theta$ - Horizontal acceleration: $a = 0$ (no forces acting horizontally) - Horizontal velocity remains constant throughout flight ### Range Calculation - Since horizontal motion has constant velocity: $$\text{Range} = \text{horizontal velocity} \times \text{time of flight}$$ $$R = (u \cos \theta) \times \frac{2u \sin \theta}{g}$$ - Simplify using the trigonometric identity $2 \sin \theta \cos \theta = \sin 2\theta$: $$R = \frac{u^2 \sin 2\theta}{g}$$ - Maximum range occurs when $\theta = 45°$ (since $\sin 90° = 1$ is maximum) ### How Range Varies with Launch Angle For a fixed launch speed of 20 m s⁻¹, the range varies with angle:

- Notice the symmetry: 30° and 60° give the same range, as do 15° and 75° - The maximum range occurs at 45° as predicted by the $\sin 2\theta$ relationship ## Three Scenarios for Projectile Motion ### 1. Vertical Projection (Free Fall) - Object dropped from rest or thrown straight up/down - Only vertical motion exists (no horizontal component) - Initial conditions: $u_x = 0$, acceleration $a = g$ (downward) - Example: Dropping a stone from a bridge or throwing a ball straight up - Relevant SUVAT equations: - $v = u + gt$ - $s = ut + \frac{1}{2}gt^2$ - $v^2 = u^2 + 2gs$ ### 2. Horizontal Projection - Object launched horizontally from a height - Initial vertical velocity: $u_y = 0$ - Initial horizontal velocity: $u_x = u$ (constant) - Example: Ball rolling off a table edge, plane dropping supplies **Solution approach:** 1. Use vertical motion to find time: $h = \frac{1}{2}gt^2$ → $t = \sqrt{\frac{2h}{g}}$ 2. Use this time to find horizontal range: $R = u_x t$ ### 3. Projection at an Angle - Most general case with both components present from launch - Requires resolving initial velocity into components: - Vertical: $u_y = u \sin \theta$ - Horizontal: $u_x = u \cos \theta$ - Examples: Kicking a football, firing a projectile, water from a fountain #### Trajectory Shape The path of a projectile launched at an angle forms a parabola. For a projectile launched at 20 m s⁻¹ at 60°:

- The parabolic shape is characteristic of all projectile motion (with air resistance ignored) - Peak occurs when vertical velocity reaches zero - Trajectory is symmetric for launch and landing at same height

Always remember that vertical acceleration $a = -g$ is negative when upward is positive

## Problem-Solving Strategy ### Step 1: Identify the Scenario - Determine if it's vertical, horizontal, or angled projection - Sketch a diagram showing the trajectory and given information ### Step 2: List Known Quantities **For vertical motion:** - Displacement: $s_y$ or $h$ - Initial velocity: $u_y$ (may need resolving from angle) - Final velocity: $v_y$ - Acceleration: $a = \pm g$ (sign depends on chosen direction) - Time: $t$ **For horizontal motion:** - Displacement: $s_x$ or $R$ - Velocity: $u_x = v_x$ (constant throughout) - Acceleration: $a = 0$ - Time: $t$ (same as vertical time) ### Step 3: Select Appropriate Equations **Vertical motion** (with $a = g$): - $v = u + gt$ - $s = ut + \frac{1}{2}gt^2$ - $v^2 = u^2 + 2gs$ **Horizontal motion** (with $a = 0$): - Distance = speed × time - $R = u_x t$ ### Step 4: Solve and Check Units - Work through the algebra carefully - Ensure all values are in SI units (metres, seconds) - Check the answer makes physical sense ## Example Problem Types ### Type 1: Finding Time in Flight (Vertical Free Fall) - Given: height $h$ and initial vertical velocity $u = 0$ - Find: time $t$ - Equation: $h = \frac{1}{2}gt^2$ - Solution: $t = \sqrt{\frac{2h}{g}}$ - Example: A feather dropped in vacuum tube from height 1.25 m takes approximately 0.5 s ### Type 2: Finding Launch Speed (Horizontal Projection) - Given: height $h$ and horizontal range $R$ - Find: initial horizontal speed $u$ - First find time from vertical motion: $t = \sqrt{\frac{2h}{g}}$ - Then use horizontal motion: $u = \frac{R}{t}$ - Example: Motorcycle jumping 10 m horizontally from 1.25 m height requires launch speed ≈ 20 m s⁻¹ ### Type 3: Finding Maximum Height (Angled Projection) - Given: initial speed $u$ and angle $\theta$ - Find: maximum height $H$ - Resolve vertical component: $u_y = u \sin \theta$ - At maximum height, $v = 0$ - Use: $v^2 = u_y^2 - 2gH$ - Solution: $H = \frac{(u \sin \theta)^2}{2g}$ - Example: Ball thrown at 12 m s⁻¹ at 50° reaches height ≈ 4.3 m ## Common Mistakes to Avoid - Forgetting that acceleration is **negative** ($-g$) when the object is rising (if upward is positive) - Confusing which component uses $\sin \theta$ and which uses $\cos \theta$: - Vertical component → $\sin \theta$ (remember: **s**in for **s**ky) - Horizontal component → $\cos \theta$ - Not converting all measurements to SI units (metres and seconds) before calculating - Using the wrong time value – vertical and horizontal motions share the same time of flight - Forgetting that horizontal velocity remains constant (common error is to apply acceleration horizontally) ## Key Relationships Summary | Parameter | Proportional To | Notes | |-----------|----------------|-------| | Time of flight | $u \sin \theta$ | Linear relationship with vertical component | | Maximum height | $(u \sin \theta)^2$ | Quadratic relationship with vertical component | | Range | $u^2 \sin 2\theta$ | Maximum at 45° launch angle | | Horizontal displacement | Time only | Since horizontal velocity is constant |

For symmetric trajectories, the projectile takes equal time going up and coming down, and lands with the same speed it was launched (but opposite vertical direction)