Two-Dimensional Kinematics and Projectile Motion Study Guide

Tutoring Services: Academic assistance and tutoring are available in room C 105, located directly across the hall from the classroom.
Course Management: Students are encouraged to stay current with assignments and attend all classes. If a student falls behind due to work commitments, they must take responsibility for redoing missing material and checking their grades regularly.

Compound Motion Definition: Projectile motion is viewed as a compound motion consisting of two independent components: motion in the horizontal ( $x$ ) direction and motion in the vertical ( $y$ ) direction.
Independence of Motion: The most critical experimental fact is that horizontal and vertical motions are completely independent of each other. A change in motion in one direction does not affect the motion in the other direction.
The Baseball Metaphor: If a baseball is tossed along a path, its vertical motion behaves exactly like a ball tossed straight up and down under the influence of gravity, regardless of its horizontal speed.
Vertical Dynamics: * Gravity is the only force acting on the object (neglecting air resistance). * The object accelerates downward at a constant rate of $g = 9.8\,m/s^2$ . In a coordinate system where "up" is positive, the acceleration in the $y$ direction ( $a_y$ ) is $-9.8\,m/s^2$ . * As the object ascends, its vertical velocity ( $v_y$ ) decreases until it reaches its peak, where $v_y = 0$ . At this point, the object has reached its maximum altitude. * As it descends, its vertical velocity increases in the negative direction.
Horizontal Dynamics: * In the absence of air resistance, there are no horizontal forces acting on the projectile. * Consequently, the horizontal component of velocity ( $v_x$ ) remains constant throughout the entire flight.

The Arrow Example: * If two arrows are released at the same time—one dropped vertically and one shot horizontally—they will hit the ground at the exact same time. * This occurs because the same force of gravity is acting on both, providing the same downward acceleration of $-9.8\,m/s^2$ . The horizontal velocity of the second arrow does not change the time it takes to fall to the ground.
Launch Angles and Horizontal Range: * Complementary Angles: Objects launched at complementary angles (angles that add up to $90^{\circ}$ ) will land at the same horizontal distance (range), provided the initial velocity is the same. * Examples of Complementary Pairs: $15^{\circ}$ and $75^{\circ}$ ; $30^{\circ}$ and $60^{\circ}$ . * Maximum Range: An object launched at an angle of $45^{\circ}$ will achieve the furthest horizontal range.
The Target and Pellet Experiment: * Scenario: A gun fires a pellet at a target at the exact same instant the target is released and allowed to fall. * Result: The pellet will always collide with the target. Because gravity acts on both the pellet and the target at the same rate, they undergo the same vertical displacement over the same interval of time. * Note on Sights: Rifle sights are designed to account for this gravitational "drop" over long ranges.

Initial Velocity Components: A projectile launched with an initial velocity $v_0$ at a projection angle $\theta_0$ can be resolved into: * $v_{0x} = v_0 \times \cos(\theta_0)$ * $v_{0y} = v_0 \times \sin(\theta_0)$
Vector Magnitude and Direction at Any Point: * The magnitude of the velocity vector is given by: $v = \sqrt{v_x^2 + v_y^2}$ * The angle of the vector relative to the horizontal is given by: $\theta = \tan^{-1}\left(\frac{v_y}{v_x}\right)$
Calculating the Projection Angle: The inverse tangent function (arctan) must be used with care as it only returns values between $-90^{\circ}$ and $+90^{\circ}$ . For vectors lying in the second or third quadrant, one must add $180^{\circ}$ to the result to find the correct angle.

Horizontal ( $X$ direction): * $v_x = v_{0x} = v_0 \times \cos(\theta_0)$ (Constant) * $\Delta x = v_{0x} \times t$
Vertical ( $Y$ direction): * $a_y = -g = -9.8\,m/s^2$ * $v_y = v_0 \times \sin(\theta_0) - g \times t$ * $\Delta y = (v_0 \times \sin(\theta_0)) \times t - \frac{1}{2} \times g \times t^2$ * $v_y^2 = (v_0 \times \sin(\theta_0))^2 - 2 \times g \times \Delta y$
Superposition: Projectile motion is the "superposition" or compounding of these independent $x$ and $y$ motions.

Definition: Relative velocity describes the velocity of an object as observed from a particular frame of reference (coordinate system).
Reference Frames: Most observations are made from a stationary frame fixed to the Earth, but moving frames (like a moving vehicle) are also used.
Vector Subtraction: The position of object A relative to object B ( $R_{AB}$ ) is found by: $\mathbf{R}_{AB} = \mathbf{R}_{A} - \mathbf{R}_{B}$ .
Highway Example: * An observer at rest on the side of the road sees a car traveling at $60\,mph$ . * An observer in a truck traveling at $50\,mph$ in the same direction sees the car traveling at a relative velocity of only $10\,mph$ ( $60 - 50 = 10$ ).
Relativisitic Constraint: These classical relative velocity equations do not apply to objects traveling at a sizable fraction of the speed of light.

Strategy for Projectiles: 1. Draw a sketch of the physical situation. 2. Resolve initial velocity into $x$ and $y$ components. 3. Treat horizontal motion as constant velocity ( $a=0$ ). 4. Treat vertical motion as constant acceleration ( $a=-g$ ).
Example 2.5 (Car Chase): * A car travels at a constant speed of $24\,m/s$ . A state trooper starts a chase one second later with a constant acceleration of $3\,m/s^2$ . * Resolution involves setting the position equations of both the car and the trooper equal to each other to find the time of intercept. * This typically results in a quadratic equation that can be solved via factoring or the quadratic formula.
Example 2.6 (Jetliner Landing): * A jetliner lands at $1.6 \times 10^2\,mph$ . * Deceleration rate: $10\,mph/s$ . * Conversion to SI units is mandatory: $1\,mph$ is approximately $0.447\,m/s$ . * Landing speed: $71.5\,m/s$ . * The problem calculates total displacement starting from touchdown to full stop by finding braking distance using kinematic formulas.
Example 3.6 (Ball Release): * A ball is released from a height of $45\,m$ with an initial speed of $20\,m/s$ . * Setting the origin at the point of release, the final displacement is $\Delta y = -45\,m$ . * The kinematics equations are used to solve for the time ( $t$ ) it takes for the ball to hit the ground.