Comprehensive Physics Notes: Projectile and Relative Motion

Principles of Projectile Motion

Projectile motion is a form of motion experienced by an object or particle (a projectile) that is thrown near the Earth's surface and moves along a curved path under the action of gravity only.

Acceleration: The acceleration is constant and directed downwards, represented as $a = g \downarrow$ . Specifically, the acceleration components are:
- $a_x = 0$ (no acceleration in the horizontal direction).
- $a_y = -g$ (acceleration due to gravity acting vertically downward).
Velocity Components:
- Horizontal Component ( $u_x$ ): Remains constant throughout the flight because there is no horizontal acceleration ( $a_x = 0$ ). Thus, $u_x = u \cos(\theta)$ .
- Vertical Component ( $u_y$ ): The y-component of velocity first decreases to zero at the highest point of the trajectory and then increases in the downward direction. Initial vertical velocity is $u_y = u \sin(\theta)$ .
Velocity Vector: The velocity at any time can be expressed as $\mathbf{u} = u \cos(\theta) \hat{i} + u \sin(\theta) \hat{j}$ .

Terms Associated with Projectile Motion

There are three primary parameters used to describe the motion of a projectile:

Time of Flight ( $T$ ):
- Defined as the total time during which the particle remains in the air.
- It is a property of the y-component of motion.
- Formula: $T = \frac{2u_y}{g} = \frac{2u \sin(\theta)}{g}$ .
- Maximum Time of Flight ( $T_{max}$ ) occurs when $\theta = 90^\circ$ , where $T_{max} = \frac{2u}{g}$ .
Maximum Height ( $H_{max}$ ):
- The maximum vertical height achieved by a particle from the plane of projection.
- It is also a property of the y-component.
- Formula: $H_{max} = \frac{u_y^2}{2g} = \frac{u^2 \sin^2(\theta)}{2g}$ .
- The highest possible value for $H_{max}$ occurs when $\theta = 90^\circ$ , where $H_{max} = \frac{u^2}{2g}$ .
Horizontal Range ( $R$ ):
- The total distance traveled horizontally along the plane of projection.
- Formula: $R = u_x \times T = u \cos(\theta) \times \frac{2u \sin(\theta)}{g} = \frac{2u^2 \cos(\theta) \sin(\theta)}{g} = \frac{u^2 \sin(2\theta)}{g}$ .
- Trigonometric Identity used: $\sin(2\theta) = 2 \sin(\theta) \cos(\theta)$ .
- Maximum Range ( $R_{max}$ ) occurs when $\sin(2\theta) = 1$ , which means $2\theta = 90^\circ$ or $\theta = 45^\circ$ . Thus, $R_{max} = \frac{u^2}{g}$ .

Fundamental Mathematical Relations

Relation between Maximum Height ( $H$ ) and Time of Flight ( $T$ ):
- $H = \frac{1}{g} \left(\frac{gT}{2}\right)^2 \times \frac{1}{2} = \frac{gT^2}{8}$ .
Relation between Range ( $R$ ) and Maximum Height ( $H$ ):
- Using the definitions $R = \frac{2u^2 \sin(\theta) \cos(\theta)}{g}$ and $H = \frac{u^2 \sin^2(\theta)}{2g}$ , we find:
- $\frac{H}{R} = \frac{\frac{u^2 \sin^2(\theta)}{2g}}{\frac{2u^2 \sin(\theta) \cos(\theta)}{g}} = \frac{\tan(\theta)}{4}$ .
- Simplified: $\tan(\theta) = \frac{4H}{R}$ .
Complementary Angles for Range:
- If two projectiles are fired with the same initial speed $u$ at complementary angles (angles that sum to $90^\circ$ , such as $\theta$ and $90^\circ - \theta$ ), their Horizontal Ranges will be equal ( $R_1 = R_2$ ), but their Maximum Heights will be different.

Equation of Trajectory

Trajectory motion refers to the relationship between the x and y coordinates of displacement, defining the path followed by the particle.

Standard Trajectory Equation: $y = x \tan(\theta) - \frac{gx^2}{2u^2 \cos^2(\theta)}$ .
Alternative Form: $y = x \tan(\theta) \left[ 1 - \frac{x}{R} \right]$ .
General Quadratic Form: Given as $y = ax - bx^2$ .
- Here, $a = \tan(\theta)$ , so $\theta = \tan^{-1}(a)$ .
- The range $R = \frac{a}{b}$ .
- The maximum height $H = \frac{a^2}{4b}$ .

Projectile Motion on an Inclined Plane

When a projectile is thrown on a plane inclined at an angle $\alpha$ to the horizontal:

Let the x-axis be along the inclined plane and the y-axis be perpendicular to it.
Gravity components:
- $g_x = -g \sin(\alpha)$ .
- $g_y = -g \cos(\alpha)$ .
Initial velocity components (where projection angle relative to the incline is $\beta$ ):
- $u_x = u \cos(\beta)$ .
- $u_y = u \sin(\beta)$ .
Time of Flight ( $T$ ): Found by setting displacement $y = 0$ .
- $T = \frac{2u \sin(\beta)}{g \cos(\alpha)}$ .
Range ( $R$ ) along the incline:
- $R = u \cos(\beta)T - \frac{1}{2}g \sin(\alpha)T^2$ .
- $R = \frac{2u^2 \sin(\beta) \cos(\beta + \alpha)}{g \cos^2(\alpha)}$ .

Horizontal Projectile Motion

For a particle projected horizontally with speed $u$ from a height $h$ :

Time taken to reach the ground: $T = \sqrt{\frac{2h}{g}}$ .
Horizontal Range: $R = uT = u \sqrt{\frac{2h}{g}}$ .
Velocity at any time $t$ : $\mathbf{v} = u \hat{i} - gt \hat{j}$ .
- Magnitude: $v = \sqrt{u^2 + (gt)^2}$ .
- Direction: $\tan(\phi) = \frac{gt}{u}$ .

Principles of Relative Motion

If no frame is specified, the ground is assumed to be the reference frame.

Relative Velocity: The velocity of object B with respect to object A is written as:
- $\mathbf{v}_{BA} = \mathbf{v}_B - \mathbf{v}_A$ .
Observation Principle: To observe motion from object A's perspective, imagine sitting on A (bringing it to rest) and adding the negative of A's velocity to object B.
Relative Acceleration: $\mathbf{a}_{BA} = \mathbf{a}_B - \mathbf{a}_A$ .
Special Scenario - Overtaking vs. Crossing:
- Overtake: Same direction ( $\mathbf{v}_{rel} = v_1 - v_2$ ).
- Crossing: Opposite direction ( $\mathbf{v}_{rel} = v_1 + v_2$ ).
- Orthogonal Motion: $90^\circ$ angle ( $v_{rel} = \sqrt{v_1^2 + v_2^2}$ ).

River-Crossing and Rain-Man Problems

River Crossing (Minimum Time):
- To cross in the shortest time, the swimmer must swim perpendicular to the river flow.
- $t_{min} = \frac{d}{v_s}$ , where $d$ is width and $v_s$ is swimmer speed.
- Drift $x = v_r \times t_{min}$ , where $v_r$ is river speed.
River Crossing (Minimum Distance/Drift):
- To cross to the point exactly opposite ( $drift = 0$ ), the horizontal component of swimmer's velocity must cancel the river's speed: $v_s \sin(\theta) = v_r$ .
- Condition: Possible only if v_s > v_r.
Upstream and Downstream:
- Downstream velocity: $v + u$ (with the flow).
- Upstream velocity: $v - u$ (against the flow).
Rain-Man Problems:
- Velocity of rain relative to man: $\mathbf{v}_{RM} = \mathbf{v}_R - \mathbf{v}_M$ .
- To find the angle to hold an umbrella, calculate the vector direction of $\mathbf{v}_{RM}$ .

Numerical Examples and Key Calculations

Example 1: Range = 3 times Maximum Height
- $\tan(\theta) = \frac{4H}{R} = \frac{4H}{3H} = \frac{4}{3}$ .
- $\theta = \tan^{-1}(4/3) = 53^\circ$ .
Example 2: Range for Angles $45-\theta$ and $45+\theta$
- Ratio of Time of Flight: $\frac{T_1}{T_2} = \frac{\sin(45-\theta)}{\sin(45+\theta)} = \frac{1 - \tan(\theta)}{1 + \tan(\theta)}$ .
- Ratio of Maximum Heights: $\frac{H_1}{H_2} = \frac{\sin^2(45-\theta)}{\sin^2(45+\theta)} = \left(\frac{1 - \tan(\theta)}{1 + \tan(\theta)}\right)^2$ .
Example 3: Projectiles with Same Speed at Different Angles
- Projectiles fired at $30^\circ$ and $60^\circ$ with speeds $40\,m/s$ and $60\,m/s$ .
- Ratio of Ranges: $\frac{R_1}{R_2} = \frac{u_1^2 \sin(60)}{u_2^2 \sin(120)} = \frac{40^2}{60^2} = \frac{1600}{3600} = \frac{4}{9}$ .
Example 4: Kinetic Energy at Highest Point
- Initial Kinetic Energy $K = \frac{1}{2}mu^2$ .
- At peak height, velocity is $u \cos(\theta)$ .
- $K_{peak} = \frac{1}{2}m(u \cos(\theta))^2 = K \cos^2(\theta)$ .
- For $\theta = 60^\circ$ , $K_{peak} = K \times (1/2)^2 = K/4$ .
Example 5: Multiple Collisions/Bouncing
- A particle dropped from height $h_0$ bounces with coefficient of restitution $e$ .
- Total Distance Travelled: $S = h_0 \left(\frac{1 + e^2}{1 - e^2}\right)$ .
- Total Time until Rest: $T = \sqrt{\frac{2h_0}{g}} \left(\frac{1 + e}{1 - e}\right)$ .
Example 6: Range Given Velocity Vector
- If velocity $\mathbf{u} = 6 \hat{i} + 8 \hat{j}$ , then $u_x = 6$ and $u_y = 8$ .
- $R = \frac{2 u_x u_y}{g} = \frac{2 \times 6 \times 8}{10} = 9.6\,m$ .
Example 7: Area of Bullet Spread
- Bullets fired in all directions with speed $u$ .
- Spread Area = $\pi R_{max}^2 = \pi (u^2/g)^2 = \frac{\pi u^4}{g^2}$ .