# Kinematics & Projectile Motion 📐

[cite_start]Kinematics is the study of motion without considering the forces that cause it. [cite: 1]

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## **Distance vs. Displacement**

[cite_start]It's crucial to understand the difference between these two fundamental quantities. [cite: 2]

| Feature | Distance | Displacement |
| :--- | :--- | :--- |
| **Definition** | [cite_start]The actual length of the path an object follows from its initial to final position. [cite: 2] | [cite_start]The shortest distance between an object's initial and final positions. [cite: 2] |
| **Nature** | [cite_start]It is a **scalar** quantity. [cite: 2] | [cite_start]It is a **vector** quantity. [cite: 2] |
| **Value** | [cite_start]It can never be negative. [cite: 2] | [cite_start]It can be positive, negative, or zero. [cite: 2] |
| **Relation** | [cite_start]Distance is always greater than or equal to displacement (Distance ≥ Displacement). [cite: 2] | [cite_start]Displacement is always less than or equal to distance (Displacement ≤ Distance). [cite: 2] |

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## **Velocity & Acceleration**

### **Velocity**
[cite_start]Velocity is defined as the rate of change of displacement. [cite: 4]
* [cite_start]**Nature:** It is a **vector** quantity. [cite: 4]
* [cite_start]**SI Unit:** $m/s$. [cite: 5]

**1. Average Velocity ($v_{avg}$)**
[cite_start]This is the ratio of the total displacement of a body to the total time taken. [cite: 7]
[cite_start]$$v_{avg} = \frac{\text{Total displacement}}{\text{Total time taken}} = \frac{s}{t} \text{}$$ [cite: 8]
[cite_start]If a body undergoes displacements $s_1$ and $s_2$ at times $t_1$ and $t_2$ respectively, the average velocity is: [cite: 18]
[cite_start]$$v_{avg} = \frac{s_2 - s_1}{t_2 - t_1} = \frac{\Delta s}{\Delta t} \text{}$$ [cite: 19, 20]

**2. Instantaneous Velocity ($v_{inst}$)**
[cite_start]This is the velocity of a body at a specific instant in time or at a particular point in its path. [cite: 22] [cite_start]It is the limiting value of the average velocity as the time interval approaches zero. [cite: 23]
[cite_start]$$v_{inst} = \lim_{\Delta t \to 0} \frac{\Delta s}{\Delta t} = \frac{ds}{dt} \text{}$$ [cite: 25, 26]
[cite_start]In simple terms, instantaneous velocity is the **derivative of displacement with respect to time**. [cite: 27]

### **Acceleration**
[cite_start]Acceleration is defined as the rate of change of velocity. [cite: 28]
* [cite_start]**Nature:** It is a **vector** quantity. [cite: 28]
* [cite_start]**SI Unit:** $m/s^2$. [cite: 28]

**1. Average Acceleration ($a_{avg}$)**
[cite_start]This is the ratio of the total change in velocity to the total time taken. [cite: 30]
[cite_start]$$a_{avg} = \frac{\Delta v}{\Delta t} \text{}$$ [cite: 31]

**2. Instantaneous Acceleration ($a_{inst}$)**
[cite_start]This is the acceleration of a body at a specific instant in time. [cite: 33] [cite_start]It is the limiting value of the average acceleration. [cite: 34]
[cite_start]$$a_{inst} = \lim_{\Delta t \to 0} \frac{\Delta v}{\Delta t} = \frac{dv}{dt} \text{}$$ [cite: 39]

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## [cite_start]**Equations of Motion (For Uniform Acceleration)** [cite: 40]

These equations are fundamental for solving problems involving constant acceleration.
1.  [cite_start]$v = u + at$ [cite: 41]
2.  [cite_start]$s = ut + \frac{1}{2}at^2$ [cite: 43]
3.  [cite_start]$v^2 = u^2 + 2as$ [cite: 44]
4.  [cite_start]$s = (\frac{u+v}{2})t$ [cite: 45]

[cite_start]**Motion Under Gravity:** [cite: 49]
* [cite_start]For **upward** motion, acceleration $a = -g$. [cite: 50, 51]
* [cite_start]For **downward** motion, acceleration $a = g$. [cite: 52]

#### [cite_start]**Derivation: Distance Travelled in the $n^{th}$ Second** [cite: 53]
[cite_start]The distance a body travels during a specific second (e.g., the 5th second) is given by the following derivation. [cite: 54]
* Distance in $n^{th}$ sec = (Distance in n sec) - (Distance in n-1 sec).
* [cite_start]$S_{n^{th}} = S_n - S_{n-1}$ [cite: 55]
* Substituting the formula for distance, $s = ut + \frac{1}{2}at^2$:
    [cite_start]$S_{n^{th}} = (un + \frac{1}{2}an^2) - [u(n-1) + \frac{1}{2}a(n-1)^2]$ [cite: 56]
* Expanding the terms:
    [cite_start]$S_{n^{th}} = un + \frac{1}{2}an^2 - [un - u + \frac{1}{2}a(n^2 - 2n + 1)]$ [cite: 57]
* Distributing the negative sign and canceling terms:
    [cite_start]$S_{n^{th}} = un + \frac{1}{2}an^2 - un + u - \frac{1}{2}an^2 + an - \frac{a}{2}$ [cite: 57]
* This leaves:
    [cite_start]$S_{n^{th}} = u + a(n - \frac{1}{2})$ [cite: 58]
* Factoring out 'a' gives the final formula:
    [cite_start]$$ S_{n^{th}} = u + \frac{a}{2}(2n-1) $$ [cite: 60]

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## **Projectile Motion** 🎯

[cite_start]A projectile is any object thrown into space that moves solely under the influence of gravity. [cite: 63] [cite_start]Its motion is two-dimensional. [cite: 65] [cite_start]Examples include a kicked football or a bullet fired from a gun. [cite: 64]

### [cite_start]**Type 1: Projectile Launched from the Ground** [cite: 68]
[cite_start]Consider an object projected with an initial velocity **u** at an angle **θ** with the horizontal. [cite: 79]
* [cite_start]**Horizontal Velocity ($u_x$):** $u_x = u\cos\theta$. [cite: 81] [cite_start]This component **remains constant**. [cite: 81]
* [cite_start]**Vertical Velocity ($u_y$):** $u_y = u\sin\theta$. [cite: 82] [cite_start]This component **changes** due to gravity. [cite: 82]

#### **Derivations**

**1. [cite_start]Equation of Trajectory (Path)** [cite: 83]
[cite_start]The path followed by a projectile is its trajectory. [cite: 84]
* [cite_start]Horizontal distance in time t: $x = u_x t = (u\cos\theta)t$. [cite: 85, 87]
* [cite_start]From this, $t = \frac{x}{u\cos\theta}$. [cite: 88]
* [cite_start]Vertical distance in time t: $y = u_y t - \frac{1}{2}gt^2 = (u\sin\theta)t - \frac{1}{2}gt^2$. [cite: 92, 95]
* Substituting the expression for t into the y-equation:
    [cite_start]$y = (u\sin\theta)(\frac{x}{u\cos\theta}) - \frac{1}{2}g(\frac{x}{u\cos\theta})^2$ [cite: 98]
* This simplifies to:
    [cite_start]$$y = (\tan\theta)x - (\frac{g}{2u^2\cos^2\theta})x^2 \text{}$$ [cite: 99]
    [cite_start]This is the equation of a **parabola**, so the projectile's path is parabolic. [cite: 101]

**2. Time of Flight (T)**
[cite_start]The total time the projectile remains in the air. [cite: 103]
* [cite_start]At the highest point, the vertical component of velocity is zero ($v_y = 0$). [cite: 105]
* [cite_start]Using $v_y = u_y - gt$, we get $0 = u\sin\theta - gt$. [cite: 107]
* [cite_start]Time to reach the highest point (time of ascent), $t = \frac{u\sin\theta}{g}$. [cite: 107]
* [cite_start]Since the time of ascent equals the time of descent, the total time of flight is: [cite: 108]
    [cite_start]$$T = 2t = \frac{2u\sin\theta}{g} \text{}$$ [cite: 109, 110]

**3. Maximum Height ($h_{max}$)**
[cite_start]The greatest vertical height the projectile reaches. [cite: 112]
* [cite_start]Using $v_y^2 = u_y^2 - 2gs$, at the highest point $v_y = 0$. [cite: 113]
* [cite_start]$0 = (u\sin\theta)^2 - 2gh_{max}$. [cite: 114]
* [cite_start]$$h_{max} = \frac{u^2\sin^2\theta}{2g} \text{}$$ [cite: 116]

**4. Horizontal Range (R)**
[cite_start]The total horizontal distance covered during the time of flight. [cite: 120]
* [cite_start]$R = \text{Horizontal Velocity} \times \text{Time of Flight}$. [cite: 124]
* [cite_start]$R = (u\cos\theta) \times (\frac{2u\sin\theta}{g}) = \frac{u^2(2\sin\theta\cos\theta)}{g}$. [cite: 125]
* Using the identity $\sin2\theta = 2\sin\theta\cos\theta$:
    [cite_start]$$R = \frac{u^2\sin2\theta}{g} \text{}$$ [cite: 125]
* [cite_start]The range is **maximum** when $\sin2\theta = 1$, which occurs when $2\theta = 90^\circ$, or **$\theta = 45^\circ$**. [cite: 127, 128, 129, 132, 133]
* [cite_start]$R_{max} = \frac{u^2}{g}$. [cite: 135]

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### [cite_start]**Type 2: Projectile Launched Horizontally from a Height** [cite: 170]
[cite_start]Consider an object projected horizontally with velocity **u** from a height **h**. [cite: 171]
* [cite_start]The horizontal velocity **remains constant** ($u_x = u$). [cite: 172]
* [cite_start]The initial vertical velocity is **zero** ($u_y = 0$), but it increases due to gravity. [cite: 172]

#### **Derivations**

**1. [cite_start]Equation of Trajectory** [cite: 183]
* [cite_start]Horizontal distance: $x = ut \implies t = \frac{x}{u}$. [cite: 186, 187]
* [cite_start]Vertical distance: $y = u_y t + \frac{1}{2}gt^2 = 0 + \frac{1}{2}gt^2 = \frac{1}{2}gt^2$. [cite: 188, 190]
* [cite_start]Substitute t: [cite: 191]
    [cite_start]$$y = \frac{1}{2}g(\frac{x}{u})^2 = (\frac{g}{2u^2})x^2 \text{}$$ [cite: 192]
    [cite_start]This is also the equation of a **parabola**. [cite: 193]

**2. [cite_start]Time of Flight (T)** [cite: 194]
* [cite_start]The time it takes to fall the vertical distance h. [cite: 196]
* [cite_start]Using $h = u_y T + \frac{1}{2}gT^2$, we get $h = 0 + \frac{1}{2}gT^2$. [cite: 197, 200]
* [cite_start]$$T = \sqrt{\frac{2h}{g}} \text{}$$ [cite: 201]

**3. [cite_start]Horizontal Range (R)** [cite: 203]
* [cite_start]$R = \text{Horizontal Velocity} \times \text{Time of Flight}$. [cite: 206]
* [cite_start]$$R = uT = u\sqrt{\frac{2h}{g}} \text{}$$ [cite: 207, 208]

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### **Relative Velocity**

[cite_start]The velocity of one body with respect to another is called relative velocity. [cite: 227] [cite_start]To find the relative velocity of body B with respect to body A ($\vec{V}_{BA}$), you reverse the velocity vector of A ($-\vec{V}_A$) and add it to the velocity vector of B ($\vec{V}_B$) using vector addition laws. [cite: 238, 239]
[cite_start]$$\vec{V}_{BA} = \vec{V}_B + (-\vec{V}_A) = \vec{V}_B - \vec{V}_A \text{}$$ [cite: 234]

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### [cite_start]**Conceptual Questions: A Quick Review** [cite: 241]

* **Why is velocity horizontal at the highest point?**
    [cite_start]Because the vertical component of velocity becomes zero at the highest point of the trajectory. [cite: 257]
* **At what point are velocity and acceleration perpendicular?**
    At the highest point, the velocity is purely horizontal, while acceleration (due to gravity) is purely vertical. [cite_start]Thus, they are perpendicular to each other. [cite: 260]
* **What happens to the range if the projection velocity is doubled?**
    [cite_start]Since $R \propto u^2$, doubling the velocity ($u' = 2u$) makes the new range $R' = \frac{(2u)^2\sin2\theta}{g} = 4 \times (\frac{u^2\sin2\theta}{g}) = 4R$. [cite: 263, 264, 265, 266] [cite_start]The horizontal range increases by 4 times. [cite: 269]
* **Two bullets are fired horizontally from the same height with different velocities. Which one reaches the ground first?**
    [cite_start]They both reach the ground at the same time. [cite: 273] [cite_start]The time of flight for a horizontal projectile is $T = \sqrt{\frac{2h}{g}}$, which depends only on the height 'h', not the horizontal velocity. [cite: 274]