Lesson Notes By Weeks and Term v4 - SHS 2
KINEMATICS
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KINEMATICS
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Subject: Physics
Class: SHS 2
Term: 1st Term
Week: 9
Grade code: 2.1.3.LI.2
Strand code: 1
Sub-strand code: 3
Content standard code: 2.1.3.CS.1
Indicator code: 2.1.3.LI.2
Theme: MECHANICS AND MATTER
Subtheme: KINEMATICS
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This lesson explores projectile motion, which describes the path of any object thrown or launched into the air. From a footballer like Mohammed Kudus taking a free-kick, to a market woman tossing a tuber of yam, or even water spraying from a hose, projectile motion is all around us in Ghana. Understanding the principles behind it helps us predict where an object will land and how high it will go. We will use the familiar equations of motion to analyze this two-dimensional movement by breaking it down into simpler horizontal and vertical parts.
What is a Projectile? A projectile is any object that is thrown, kicked, or otherwise launched into the air and then moves under the sole influence of gravity. We make a crucial simplifying assumption: we ignore air resistance.
The path followed by a projectile is a curve called a parabola. The Core Principle: Separation of Motion The trick to understanding projectile motion is to split the single, curved motion into two separate, simpler, and independent motions: Horizontal Motion (x-direction): There is no force acting horizontally (since we ignore air resistance). This means there is no horizontal acceleration (`a_x = 0`). The horizontal velocity is therefore constant. Vertical Motion (y-direction): Gravity is the only force acting, and it pulls the object downwards. This means there is a constant downward acceleration equal to `g` (acceleration due to gravity, approximately `9.8 m/s²` or `10 m/s²` for easier calculations). The vertical velocity is therefore constantly changing. Resolving Initial Velocity Imagine a football is kicked with an initial velocity, `u`, at an angle `θ` to the horizontal ground.
This initial velocity has two parts (components): The horizontal component: `u_x = u cos(θ)` The vertical component: `u_y = u sin(θ)`
Throughout the flight: The horizontal velocity remains `u_x`. The vertical velocity, `v_y`, changes due to gravity. Deriving the Key Formulas We will use the standard equations of motion, applying them separately to the vertical motion. (Recall: `v = u + at`, `s = ut + ½at²`, `v² = u² + 2as`) For our vertical motion: `a = -g` (negative because gravity acts downwards). Time to Reach Maximum Height (`t_up`) Concept: At the very peak of its flight (the maximum height), the projectile stops moving upwards for an instant before it starts to fall. At this point, its vertical velocity (`v_y`) is zero. Derivation: We use the equation: `v_y = u_y + a_y*t` At maximum height, `v_y = 0`. The initial vertical velocity is `u_y = u sin(θ)`. The vertical acceleration is `a_y = -g`. Substituting these values: `0 = u sin(θ) - g*t_up` Rearranging to find `t_up`: `g*t_up = u sin(θ)` Therefore, the time to reach maximum height is: > `t_up = (u sin(θ)) / g` Maximum Height (`H`) Concept: This is the maximum vertical displacement (`s_y`) reached by the projectile. We can find it using the time we just calculated, or by using the equation that doesn't involve time. Derivation: We use the equation: `v_y² = u_y² + 2*a_y*s_y` At maximum height, `v_y = 0` and the vertical displacement `s_y = H`. Substituting: `0² = (u sin(θ))² + 2(-g)H` `0 = u²sin²(θ) - 2gH` Rearranging to find `H`: `2gH = u²sin²(θ)` Therefore, the maximum height is: > `H = (u²sin²(θ)) / (2g)` Total Time of Flight (`T`) Concept: Assuming the projectile lands at the same level it was launched from, the path is symmetrical. The time it takes to go up (`t_up`) is equal to the time it takes to come down (`t_down`). Derivation: Total Time of Flight `T = t_up + t_down` Since `t_up = t_down`, we have `T = 2 * t_up` Substituting our formula for `t_up`: > `T = (2u sin(θ)) / g` Horizontal Range (`R`) Concept: The range is the total horizontal distance the projectile travels. Since horizontal velocity is constant, we use the simple formula: `distance = speed × time`. Derivation: `R = Horizontal Velocity × Total Time of Flight` `R = u_x × T` Substituting the expressions for `u_x` and `T`: `R = (u cos(θ)) × ((2u sin(θ)) / g)` `R = (2u²sin(θ)cos(θ)) / g` Using the trigonometric identity `sin(2θ) = 2sin(θ)cos(θ)`, we get the simplified formula: > `R = (u²sin(2θ)) / g`