Lesson Notes By Weeks and Term v5 - Grade 12

Lesson Video

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Performance objectives

• Define vertical projectile motion.
• Apply kinematic equations to solve problems.
• Calculate time of flight, maximum height, and final velocity.
• Draw and interpret graphs of motion.
• Explain the effect of air resistance.

Lesson summary

Projectile motion is a fundamental concept in physics that describes the motion of an object thrown or projected into the air, subject only to the acceleration of gravity. Understanding projectile motion is crucial for many real-world applications, from sports like cricket and soccer to the design of rockets and the analysis of car accidents. In the South African context, projectile motion is relevant in understanding the trajectory of a cricket ball during a match, the flight path of a soccer ball kicked towards the goal, or even the motion of water spurting from a fountain in a public park.

Lesson notes

2. 1. What is Projectile Motion? Projectile motion is the motion of an object thrown or projected into the air, subject only to the acceleration of gravity. In this week's focus, we are considering ONLY vertical projectile motion, meaning the object is launched straight up or dropped straight down. This simplifies the analysis, as we only need to consider motion in one dimension (the vertical or y-axis).

Key Assumptions: Neglecting Air Resistance: For simplicity, we initially ignore air resistance. This means that the only force acting on the projectile is gravity.

Constant Gravitational Acceleration: We assume that the acceleration due to gravity, g, is constant and directed downwards. Its value is approximately 9.8 m/s². The sign convention is critical. We will usually take 'up' as positive and 'down' as negative, meaning g = -9.8 m/s². 2.

2. Key Definitions and Concepts: Displacement (Δy): The change in position of the projectile. It's a vector quantity, meaning it has both magnitude and direction.

Velocity (v): The rate of change of displacement. It's also a vector quantity. Initial velocity (v i ) is the velocity at the start of the motion, and final velocity (v f ) is the velocity at the end of the motion.

Acceleration (a): The rate of change of velocity. In projectile motion (ignoring air resistance), the acceleration is constant and equal to the acceleration due to gravity (g).

Time (t): The duration of the motion.

Time of Flight (T): The total time the projectile spends in the air.

Maximum Height (H): The highest point reached by the projectile above its initial position. At the maximum height, the vertical velocity of the projectile is momentarily zero (v f = 0). 2.

3. Kinematic Equations for Vertical Projectile Motion: We use the following kinematic equations (derived from the definitions of displacement, velocity, and acceleration) to analyze vertical projectile motion: v f = v i + aΔt Δy = v i Δt + ½aΔt² v f ² = v i ² + 2aΔy Δy = (v i + v f )/2 * Δt Where: v f is the final velocity v i is the initial velocity a is the acceleration (in this case, g = -9.8 m/s²) Δt is the change in time Δy is the displacement Important

Note: Consistent sign conventions are crucial. If you define upward direction as positive, then g will be negative (-9.8 m/s²). 2.

4. Worked

Examples: Example 1: A ball is thrown vertically upwards from the ground with an initial velocity of 15 m/s. Neglecting air resistance, calculate: a) The maximum height reached by the ball. b) The time it takes for the ball to reach its maximum height. c) The velocity of the ball when it returns to the ground.

Solution: a) To find the maximum height (Δy), we know that v f = 0 m/s at the maximum height. We also know v i = 15 m/s and a = -9.8 m/s². Using the equation v f ² = v i ² + 2aΔy: 0² = 15² + 2(-9.8)Δy 0 = 225 - 19.6Δy 19.6Δy = 225 Δy = 225 / 19.6 Δy = 11.48 m Therefore, the maximum height reached by the ball is 11.48 meters. b) To find the time (Δt) to reach the maximum height, we can use the equation v f = v i + aΔt: 0 = 15 + (-9.8)Δt 8Δt = 15 Δt = 15 / 9.8 Δt = 1.53 s Therefore, it takes 1.53 seconds for the ball to reach its maximum height. c) When the ball returns to the ground, its displacement Δy = 0 m (relative to the starting point). We can use v f ² = v i ² + 2aΔy again: v f ² = 15² + 2(-9.8)(0) v f ² = 225 v f = ±√225 v f = ±15 m/s Since the ball is moving downwards, we take the negative root: v f = -15 m/s. This makes sense; in the absence of air resistance, the final velocity is equal in magnitude but opposite in direction to the initial velocity.

Example 2: A stone is dropped from the top of a building 20 m high.

Calculate: a) The time it takes for the stone to reach the ground. b) The velocity of the stone just before it hits the ground.

Solution: a) In this case, v i = 0 m/s (since the stone is dropped), Δy = -20 m (downwards displacement), and a = -9.8 m/s². We can use the equation Δy = v i Δt + ½aΔt²: -20 = 0*Δt + ½(-9.8)Δt² -20 = -4.9Δt² Δt² = -20 / -4.9 Δt² = 4.08 Δt = √4.08 Δt = 2.02 s Therefore, it takes 2.02 seconds for the stone to reach the ground. b) To find the final velocity (v f ), we can use the equation v f = v i + aΔt: v f = 0 + (-9.8)(2.02) v f = -19.8 m/s Therefore, the velocity of the stone just before it hits the ground is -19.8 m/s (downwards). 2.

5. Graphs of Vertical Projectile Motion: Displacement-Time Graph (Δy vs. t): A parabola. If the object is thrown upwards, the parabola opens downwards. The slope of the tangent at any point gives the instantaneous velocity at that time. Velocity-Time Graph (v vs. t): A straight line with a negative slope (since acceleration is constant and negative, due to gravity if up is positive). The slope of the line represents the acceleration due to gravity. The area under the graph represents the displacement. Acceleration-Time Graph (a vs. t): A horizontal straight line at a = -9.8 m/s² (assuming upward as positive). 2.6.