Lesson Notes By Weeks and Term v4 - SHS 2

Lesson Video

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Performance objectives

• Define projectile motion and its range.
• Calculate the range of a projectile using the correct formula.
• Explain the factors that affect the range of a projectile.
• Solve problems involving projectile motion.

Lesson summary

Today, we are exploring one of the most exciting aspects of motion: projectile motion. Think about a footballer, like Mohammed Kudus, taking a free kick. He has to decide how hard to kick the ball and at what angle to score a goal or pass to a teammate. The path the ball follows through the air is a projectile path. Understanding this physics helps us predict where the ball will land. This is not just for sports; it applies to everything from a stone thrown into the Volta Lake to water spraying from an irrigator on a farm in the Afram Plains. Our main goal is to learn how to calculate the range—the total horizontal distance a projectile travels before it lands.

Lesson notes

This section breaks down the physics needed to determine the range of a projectile. We will build our understanding step-by-step. A. What is a Projectile? A projectile is any object that is thrown or projected into the air and then moves under the influence of gravity alone. We make a key assumption for our calculations: we ignore air resistance. Examples: A kicked football, a thrown javelin, a stone thrown from a catapult, a bullet after it leaves the gun. Path: The curved path that a projectile follows is called its trajectory, which is a parabola. B. The Most Important Principle: Independence of Motion To analyse projectile motion, we treat the horizontal and vertical motions separately. This is the secret to solving all projectile problems. Horizontal Motion (x-direction): Gravity only acts downwards, not sideways. Therefore, there is no acceleration in the horizontal direction (assuming no air resistance). The horizontal velocity of the projectile is constant throughout its flight. The equation is simple: `Distance = Speed × Time` Vertical Motion (y-direction): Gravity acts downwards. Therefore, there is a constant downward acceleration, which is the acceleration due to gravity, g (we will use g ≈ 9.8 m/s² or 10 m/s² as specified in the problem). The vertical velocity changes continuously: it decreases as the object goes up, becomes zero at the maximum height, and increases as the object comes down. We use the standard equations of motion: `v = u + at` `s = ut + ½at²` `v² = u² + 2as` C. Deriving the Formula for Range (R)

Let's consider a projectile launched from the ground with an initial velocity u at an angle θ to the horizontal.

Step 1: Resolve the Initial Velocity First, we break the initial velocity `u` into its horizontal (`u_x`) and vertical (`u_y`) components using trigonometry (SOH CAH TOA). Horizontal component: `u_x = u cos(θ)` (This velocity remains constant) Vertical component: `u_y = u sin(θ)` (This velocity is affected by gravity)

Step 2: Find the Time of Flight (T) The time of flight is the total time the projectile spends in the air. The projectile lands back at the same vertical level it started from, so its total vertical displacement (`s_y`) is zero. We use the equation: `s_y = u_y*t + ½a_y*t²` Knowns for vertical motion: `s_y = 0` (it lands back on the ground) `u_y = u sin(θ)` `a_y = -g` (negative because gravity acts downwards) Substitute these into the equation: `0 = (u sin(θ))t - ½gt²` We can factor out `t`: `t(u sin(θ) - ½gt) = 0` This gives two solutions for `t`: `t = 0` (This is the start of the motion, when it's on the ground) `u sin(θ) - ½gt = 0` => `½gt = u sin(θ)` => `t = (2u sin(θ)) / g`

Evaluation guide

• Determine the range of a projectile.
• Project – based learning : Let learners determine the range as the horizontal distance travelled by
• the projectile.
• conceptual