Comprehensive Study Notes: 1D & 2D Motion, Kinematics, Vectors, and Projectiles

Test Instructions and General Guidelines

• Structure and Preparation Strategy:

  • Read and master foundational topics in sequential order prior to attempting assessments.

  • Complete practice problems independently before reviewing provided solutions.

  • Always provide correct physical units with every numerical answer.

  • Testing formats may include both free-response problems and multiple-choice questions.

  • Cumulative content: Topics from Test 1 remain examinable and must be reviewed separately.

• Gravitational Acceleration Conventions:

  • On official test examinations, use the standardized approximation $g = 10\,\text{m/s}^2$.

  • Curricular review sheets and standard textbook problems frequently utilize $g = 9.8\,\text{m/s}^2$ (or $9.80\,\text{m/s}^2$), which accounts for minor numerical deviations between calculated solutions and review answer keys.

Fundamentals of SI Units and Physical Quantities

• International System of Units (SI):

  • The SI system provides the universal metric foundation for scientific measurement.

  • Fundamental (Base) Units: Inherent, indivisible physical quantities serving as building blocks for all other units:

  • Length: meter ($\text{m}$

  • Mass: kilogram ($\text{kg}$

  • Time: second ($\text{s}$

  • Thermodynamic temperature: kelvin ($\text{K}$

  • Electric current: ampere ($\text{A}$

  • Amount of substance: mole ($\text{mol}$

  • Luminous intensity: candela ($\text{cd}$

  • Derived Units: Combinations constructed from base units via physical laws and mathematical relations:

  • Velocity / Speed: $\text{m/s}$ ($\text{m}\cdot\text{s}^{-1}$

  • Acceleration: $\text{m/s}^2$ ($\text{m}\cdot\text{s}^{-2}$

  • Force: newton ($\text{N}$), defined as $\text{kg}\cdot\text{m/s}^2$

  • Energy / Work: joule ($\text{J}$), defined as $\text{kg}\cdot\text{m}^2/\text{s}^2$

  • Area: $\text{m}^2$

  • Volume: $\text{m}^3$

• Standard Metric Multipliers and Prefixes:

  • Giga ($\text{G}$): $10^9$

  • Mega ($\text{M}$): $10^6$

  • Kilo ($\text{k}$): $10^3$

  • Centi ($\text{c}$): $10^{-2}$

  • Milli ($\text{m}$): $10^{-3}$

  • Micro ($\mu$): $10^{-6}$

  • Nano ($\text{n}$): $10^{-9}$

  • Practical conversions:

  • $1.50\,\text{ms} = 1.50 \times 10^{-3}\,\text{s}$

  • $2.5\,\text{km} = 2500\,\text{m}$

  • Conversion factor for velocity: $\text{km/h} \div 3.6 = \text{m/s}$

• Order-of-Magnitude Estimations (Lengths and Masses):

  • Reference values for physical lengths:

  • Width of a fingernail: $\approx 1\,\text{cm}$ ($0.01\,\text{m}$

  • Height of an interior doorway: $\approx 2\,\text{m}$

  • Height of an adult human: $\approx 1.7\,\text{m}$

  • Length of a typical classroom: $\approx 10\,\text{m}$

  • Length of a football field: $\approx 100\,\text{m}$

  • Reference values for physical masses:

  • Small metal paper clip: $\approx 1\,\text{g}$ ($0.001\,\text{kg}$

  • Typical apple: $\approx 0.1\,\text{kg}$ ($100\,\text{g}$

  • Textbook or one liter of water: $\approx 1\,\text{kg}$

  • Bicycle: $\approx 10\text{--}15\,\text{kg}$

  • Adult human: $\approx 60\text{--}80\,\text{kg}$

  • Passenger car: $\approx 1000\text{--}1500\,\text{kg}$

Scalars, Vectors, and Vector Operations

• Classification of Physical Quantities:

  • Scalar: A quantity completely specified by magnitude (numerical size and unit) without spatial direction. Examples: distance, speed, time, mass, temperature, energy, volume, density.

  • Vector: A quantity possessing both magnitude and spatial direction. Examples: displacement, velocity, acceleration, force, momentum.

• Distance vs. Displacement:

  • Distance ($d$): The total scalar length of the actual path traversed regardless of direction.

  • Displacement ($\Delta x$ or $\vec{d}$): The straight-line vector pointing from the initial coordinate to the final coordinate.

  • Comparative scenario: Walking $4\,\text{m}$ east followed by $4\,\text{m}$ west yields a cumulative distance of $8\,\text{m}$, but a resultant displacement of $0\,\text{m}$.

• Speed vs. Velocity:

  • Speed: Rate of change of distance over time (scalar: "how fast").

  • Velocity: Rate of change of position with respect to time in a specified direction (vector: "how fast and in what direction").

• Graphical Vector Addition (Head-to-Tail Method):

  • Step 1: Draw the initial vector to scale pointing in its stated direction.

  • Step 2: Draw the subsequent vector starting with its tail precisely positioned at the head (arrowhead) of the preceding vector.

  • Step 3: Repeat sequentially for all additional vectors.

  • Step 4: Construct the resultant vector ($\vec{R}$) as an arrow directed from the tail of the very first vector to the head of the final vector.

  • Commutative property: Vector addition is independent of arrangement order ($\vec{A} + \vec{B} = \vec{B} + \vec{A}$).

  • Finding a missing vector from a known resultant:

$\vec{A}_{\text{missing}} = \vec{R} - \sum \vec{A}_{\text{known}}$

• Vector Equilibrium and Equilibrants:

  • A state of static equilibrium requires the vector sum of all concurrent vectors to equal zero ($\vec{R} = 0$).

  • When drawn head-to-tail, vectors in equilibrium necessarily form a closed polygon whose terminus closes back on the starting origin.

  • The equilibrant is a vector equal in magnitude and exactly opposite in direction ($180^\circ$ opposite) to the resultant of the active system.

  • Classic geometric example: A closed right triangle of $3\,\text{N}$, $4\,\text{N}$, and $5\,\text{N}$ ($3^2 + 4^2 = 5^2$) closes upon itself, confirming $\vec{R} = 0$.

• Mathematical Resolution of Vectors:

  • Resolving a vector into orthogonal Cartesian components using counterclockwise angle $\theta$ measured from the positive $+x$-axis:

$A_x = A \cos(\theta)$

$A_y = A \sin(\theta)$

$\cos(\theta) = \frac{A_x}{A}$

$\sin(\theta) = \frac{A_y}{A}$

$\tan(\theta) = \frac{A_y}{A_x}$

• Directional signs: Components directed along the negative coordinate axes (leftward or downward) are defined as negative.

• Analytic summation of multiple vectors:

$R_x = \sum A_x$

$R_y = \sum A_y$

$R = \sqrt{R_x^2 + R_y^2}$

$\theta = \tan^{-1}\left(\frac{R_y}{R_x}\right)$

• Worked component example: A soccer ball kicked at $20\,\text{m/s}$ at an angle of $30^\circ$ above the horizontal:

$v_x = 20 \cos(30^\circ) = 17.3\,\text{m/s}$

$v_y = 20 \sin(30^\circ) = 10\,\text{m/s}$

• Worked perpendicular vector example: A hiker walks $6.0\,\text{m}$ east followed by $8.0\,\text{m}$ north:

$R = \sqrt{(6.0)^2 + (8.0)^2} = \sqrt{36 + 64} = 10\,\text{m}$

$\theta = \tan^{-1}\left(\frac{8.0}{6.0}\right) = 53^\circ\,\text{north of east}$

• Concurrent Vectors:

  • Vectors that act concurrently pass through the exact same spatial point.

  • The magnitude of the resultant varies exclusively with the angle between them:

  • Maximum resultant ($\theta = 0^\circ$, parallel / same direction):

$R_{\max} = A + B$

• Minimum resultant ($\theta = 180^\circ$, anti-parallel / opposite direction):

$R_{\min} = |A - B|$

• Perpendicular configuration ($\theta = 90^\circ$):

$R = \sqrt{A^2 + B^2}$

• Allowable range: Any resultant value within the closed domain $[|A - B|, A + B]$ is physically achievable depending on the orientation angle.

  • Example with forces of $5\,\text{N}$ and $8\,\text{N}$:

• $R_{\max} = 8 + 5 = 13\,\text{N}$

• $R_{\min} = 8 - 5 = 3\,\text{N}$

• Possible resultants: Any magnitude from $3\,\text{N}$ to $13\,\text{N}$. Values such as $2\,\text{N}$ or $15\,\text{N}$ are strictly impossible; $10\,\text{N}$ is physically possible.

  • Systems with three or more concurrent vectors:

• Maximum resultant equals the algebraic scalar sum of all vector magnitudes.

• Minimum resultant equals $0\,\text{N}$ if the largest individual vector is less than or equal to the sum of the remaining vectors ($A_{\max} \le \sum A_{\text{others}}$).

• If the largest vector exceeds the sum of the others, the minimum possible resultant equals $A_{\max} - \sum A_{\text{others}}$.

• Example: For $3\,\text{N}$, $4\,\text{N}$, and $5\,\text{N}$, $5 \le 3 + 4$, meaning the minimum resultant is $0\,\text{N}$.

• Example: For $3\,\text{N}$, $4\,\text{N}$, and $10\,\text{N}$, $10 > 3 + 4 = 7$, so the absolute minimum resultant is $10 - 7 = 3\,\text{N}$.

Motion Diagrams and Velocity Changes

• Visualizing Kinematics with Motion Diagrams:

  • Motion diagrams plot position coordinates at uniform, consecutive time increments ($\Delta t$).

  • Velocity vectors ($\vec{v}$) point in the direction of motion; acceleration vectors ($\vec{a}$) indicate the direction of the net velocity change.

  • Constant velocity ($a = 0$): Dots are uniformly spaced across equal time intervals.

  • Speeding up: Progressive spacing between consecutive dots expands over time; $\vec{v}$ and $\vec{a}$ point in the identical direction.

  • Slowing down: Progressive spacing between consecutive dots contracts over time; $\vec{v}$ and $\vec{a}$ point in opposing directions.

• The Direction and Sign Rule:

  • Acceleration is defined strictly by the rate of change of velocity ($\vec{a} = \frac{\Delta \vec{v}}{\Delta t}$), not by the absolute direction of motion.

  • When velocity and acceleration share the same sign (both positive or both negative), the object speeds up.

  • When velocity and acceleration possess opposite signs, the object slows down.

  • Algebraic interpretation of signs:

  • Moving in the negative direction ($v < 0$) and speeding up requires a negative acceleration ($a < 0$).

  • Moving in the negative direction ($v < 0$) and slowing down requires a positive acceleration ($a > 0$).

  • The operational direction of acceleration during deceleration is always directly opposite the instantaneous velocity vector.

• Kinematics with Directional Reversals:

  • Reversing direction mandates a formal algebraic sign change in velocity.

  • Common pitfall: Dropping the negative sign during vector velocity changes.

  • Worked impact example: A $0.02\,\text{kg}$ ball strikes the floor moving downward at $2.5\,\text{m/s}$ and rebounds upward at $2.2\,\text{m/s}$, with an impact contact duration of $1.50\,\text{ms}$. Taking upward as positive:

$v_i = -2.5\,\text{m/s}$

$v_f = +2.2\,\text{m/s}$

$\Delta v = v_f - v_i = 2.2 - (-2.5) = +4.7\,\text{m/s}$

$a = \frac{\Delta v}{\Delta t} = \frac{4.7}{1.50 \times 10^{-3}} = 3.1 \times 10^3\,\text{m/s}^2\,\text{(directed upward)}$

• Notice that the object's mass ($0.02\,\text{kg}$) is superfluous to computing acceleration.

Graphical Analysis of Motion (x-t, v-t, and a-t Graphs)

• Calculus Foundations via Slope and Area:

  • Position vs. Time ($x\text{--}t$):

  • Slope ($\frac{\Delta x}{\Delta t}$) represents instantaneous/average velocity ($v$).

  • Area under the curve possesses no direct physical meaning.

  • Velocity vs. Time ($v\text{--}t$):

  • Slope ($\frac{\Delta v}{\Delta t}$) represents acceleration ($a$).

  • Area under the curve ($\int v\,dt$ or base $\times$ height) represents displacement ($\Delta x$).

  • Acceleration vs. Time ($a\text{--}t$):

  • Slope possesses no elementary physical application here.

  • Area under the curve ($\int a\,dt$ or base $\times$ height) represents change in velocity ($\Delta v$).

  • Area calculations below the horizontal time axis reflect negative displacement or negative velocity changes.

• Geometric Profiles and Physical Interpretations:

  • Position vs. Time ($x\text{--}t$):

  • Flat horizontal line: Static equilibrium, object at rest ($v = 0$).

  • Straight sloped line: Uniform constant velocity ($a = 0$); steeper slopes denote higher speed.

  • Parabolic curve growing steeper upward: Positive acceleration (speeding up in the positive direction).

  • Downward-sloping straight line: Motion in the negative coordinate direction with constant velocity.

  • Velocity vs. Time ($v\text{--}t$):

  • Flat horizontal line: Zero acceleration ($a = 0$), constant steady velocity.

  • Straight linear sloped line: Constant, uniform acceleration ($a = \text{constant}$; kinematics apply).

  • Line intersecting the horizontal time axis ($v = 0$): Instantaneous stop and immediate direction reversal.

  • Acceleration vs. Time ($a\text{--}t$):

  • Flat horizontal line off zero: Constant non-zero acceleration.

  • Flat horizontal line at zero: Constant velocity or static rest.

• Comprehensive Worked Example for a Velocity-Time Profile:

  • Scenario: A cart accelerates from $0$ to $10\,\text{m/s}$ uniformly over $5\,\text{s}$, maintains $10\,\text{m/s}$ for $3\,\text{s}$, and decelerates uniformly to rest in $2\,\text{s}$.

  • Accelerations via piecewise slopes:

$a_1 = \frac{10 - 0}{5} = +2\,\text{m/s}^2$

$a_2 = \frac{10 - 10}{3} = 0\,\text{m/s}^2$

$a_3 = \frac{0 - 10}{2} = -5\,\text{m/s}^2$

• Displacements via piecewise geometric areas:

$\Delta x_1 = \frac{1}{2}(5)(10) = 25\,\text{m}$

$\Delta x_2 = (3)(10) = 30\,\text{m}$

$\Delta x_3 = \frac{1}{2}(2)(10) = 10\,\text{m}$

$\Delta x_{\text{total}} = 25 + 30 + 10 = 65\,\text{m}$

• Worked Example for a Position-Time Profile:

  • Scenario: An object's position changes linearly from $0\,\text{m}$ at $t = 0\,\text{s}$ to $20\,\text{m}$ at $t = 4\,\text{s}$.

  • Slope computation:

$v = \frac{20 - 0}{4 - 0} = 5\,\text{m/s}$

• Constant slope confirms constant velocity; therefore acceleration $a = 0\,\text{m/s}^2$.

Kinematic Equations for Uniform Acceleration

• Governing Formulas:

  • These relations are strictly valid only when acceleration is constant throughout the motion:

$v_f = v_i + at$

$d = v_i t + \frac{1}{2}at^2$

$v_f^2 = v_i^2 + 2ad$

• Supplementary and Specialized Relations:

$a = \frac{v_f - v_i}{t}$

$v_{\text{avg}} = \frac{v_i + v_f}{2}$

$d = \left(\frac{v_i + v_f}{2}\right)t$

• Special Case for Motion from Rest ($v_i = 0$):

$d = \frac{1}{2}at^2$

$a = \frac{2d}{t^2}$

• Equation Selection Matrix (The "Missing Variable" Strategy):

  • If displacement ($d$) is neither provided nor required: select $v_f = v_i + at$

  • If final velocity ($v_f$) is neither provided nor required: select $d = v_i t + \frac{1}{2}at^2$

  • If elapsed time ($t$) is neither provided nor required: select $v_f^2 = v_i^2 + 2ad$

• Algorithmic Problem-Solving Protocol:

  • Step 1: Sketch the physical layout and designate an explicit positive spatial direction.

  • Step 2: Construct a comprehensive parameter table listing $d$, $a$, $t$, $v_i$, and $v_f$. Identify explicit verbal cues ("starts from rest" specifies $v_i = 0$; "comes to a stop" specifies $v_f = 0$).

  • Step 3: Identify the single unmentioned parameter to pinpoint the necessary formula.

  • Step 4: Isolate the unknown symbolically, compute, assign appropriate units, and check the sign.

• Comprehensive Worked Kinematic Problems:

  • Braking Vehicle: A vehicle traveling at $20\,\text{m/s}$ experiences an acceleration of $-4\,\text{m/s}^2$ until brought to rest:

$t = \frac{v_f - v_i}{a} = \frac{0 - 20}{-4} = 5\,\text{s}$

$d = \frac{v_f^2 - v_i^2}{2a} = \frac{0 - (20)^2}{2(-4)} = \frac{-400}{-8} = 50\,\text{m}$

• Unit Conversion Precursor: A vehicle accelerates from rest to $60\,\text{km/h}$ in $5.00\,\text{s}$:

$v_f = 60\,\text{km/h} \div 3.6 = 16.7\,\text{m/s}$

$a = \frac{16.7 - 0}{5.00} = 3.3\,\text{m/s}^2$

• Acceleration from Rest Without Final Velocity: A skateboard accelerates from rest across $45\,\text{m}$ in $6\,\text{s}$:

$d = \frac{1}{2}at^2 \implies a = \frac{2d}{t^2} = \frac{2(45)}{6^2} = \frac{90}{36} = 2.5\,\text{m/s}^2$

• Inclined Ramp Descent: An object accelerates down an incline with $v_i = 1.0\,\text{m/s}$, $a = 0.2\,\text{m/s}^2$, for $t = 20.0\,\text{s}$:

$v_f = v_i + at = 1.0 + 0.2(20.0) = 5\,\text{m/s}$

Free Fall Dynamics and One-Dimensional Vertical Motion

• Core Physical Principles of Free Fall:

  • Free fall describes motion governed solely by the influence of gravity, ignoring aerodynamic drag.

  • Every free-falling object undergoes identical constant downward acceleration ($g = 9.8\,\text{m/s}^2$, or $10\,\text{m/s}^2$ on test assessments) irrespective of mass, shape, or instantaneous direction of motion.

  • Acceleration at the absolute peak of vertical flight is never zero; it remains exactly $g$ directed downward.

• Dropped Objects ($v_{iy} = 0$):

  • Mathematical formulas for objects released from rest:

$y = \frac{1}{2}gt^2$

$t = \sqrt{\frac{2y}{g}}$

$v_f = gt$

$v_f = \sqrt{2gy}$

• Progression with $g = 10\,\text{m/s}^2$:

  • At $t = 0\,\text{s}$, speed $v = 0\,\text{m/s}$, distance fallen $d = 0\,\text{m}$

  • At $t = 1\,\text{s}$, speed $v = 10\,\text{m/s}$, distance fallen $d = 5\,\text{m}$

  • At $t = 2\,\text{s}$, speed $v = 20\,\text{m/s}$, distance fallen $d = 20\,\text{m}$

  • At $t = 3\,\text{s}$, speed $v = 30\,\text{m/s}$, distance fallen $d = 45\,\text{m}$

  • At $t = 4\,\text{s}$, speed $v = 40\,\text{m/s}$, distance fallen $d = 80\,\text{m}$

• Worked dropped rock example: A stone dropped from rest hits the ground after $1.5\,\text{s}$ ($g = 10\,\text{m/s}^2$):

$v = gt = 10(1.5) = 15\,\text{m/s}$

$y = \frac{1}{2}(10)(1.5)^2 = 11\,\text{m}$

• Objects Thrown Downward ($v_{iy} \ne 0$):

• Initial velocity cannot be eliminated; use full kinematic forms with downward chosen as positive:

$y = v_i t + \frac{1}{2}gt^2$

• Worked examples ($g = 9.8\,\text{m/s}^2$):

  • Thrown straight down at $12\,\text{m/s}$ for $2.0\,\text{s}$:

$y = 12(2) + \frac{1}{2}(9.8)(2)^2 = 24 + 19.6 = 44\,\text{m}$

• Pushed straight down at $4.0\,\text{m/s}$ for $6\,\text{s}$:

$y = 4(6) + \frac{1}{2}(9.8)(6)^2 = 24 + 176.4 = 200\,\text{m}$

• Vertical Launches (Thrown Straight Up):

  • Symmetry of flight: Ascent and descent trajectories are strictly symmetric over level ground.

  • Total flight time is double the rise time:

$t_{\text{total}} = 2 t_{\text{up}}$

$t_{\text{up}} = \frac{v_i}{g}$

• Kinematics at peak height:

  • Instantaneous vertical velocity: $v_{\text{top}} = 0\,\text{m/s}$

  • Instantaneous acceleration: $a_{\text{top}} = g\,\text{downward}$ ($-9.8\,\text{m/s}^2$ or $-10\,\text{m/s}^2$)

• Peak height formula:

$h_{\max} = \frac{v_i^2}{2g}$

• Return speed: The object impacts the launch coordinate with a velocity equal in magnitude and opposing in direction to its launch velocity ($v_f = -v_i$).

• Worked example (Launch at $20\,\text{m/s}$, $g = 10\,\text{m/s}^2$):

$t_{\text{up}} = \frac{20}{10} = 2\,\text{s}$

$t_{\text{total}} = 2(2) = 4\,\text{s}$

$h_{\max} = \frac{(20)^2}{2(10)} = 20\,\text{m}$

• Velocity sequence across each successive second: $+20\,\text{m/s}$, $+10\,\text{m/s}$, $0\,\text{m/s}$, $-10\,\text{m/s}$, $-20\,\text{m/s}$.

  • Worked example (Caught after $8.4\,\text{s}$, $g = 10\,\text{m/s}^2$):

$t_{\text{up}} = \frac{8.4}{2} = 4.2\,\text{s}$

$v_i = g t_{\text{up}} = 10(4.2) = 42\,\text{m/s}$

• Speed at peak = $0\,\text{m/s}$; acceleration at peak = $-10\,\text{m/s}^2$; reception speed = $42\,\text{m/s}$.

  • Worked example (Determining launch velocity from height without time, $g = 9.8\,\text{m/s}^2$):

• A frog leaps vertically to a maximum altitude of $0.6\,\text{m}$:

$v_i = \sqrt{2gy} = \sqrt{2(9.8)(0.6)} = 3.4\,\text{m/s}$

• Cliff Launch Above Ground (Quadratic Method):

  • When an object is launched upward from an elevated platform of height $h$ and falls below the launch origin, set the ground coordinate as $y_f = 0$ and initial height as $y_i = h$, making net displacement negative ($\Delta y = -h$).

  • Establish upward as the positive direction ($a = -g$):

$\Delta y = v_i t - \frac{1}{2}gt^2$

$\frac{1}{2}gt^2 - v_i t + \Delta y = 0$

• Solve via the quadratic formula and reject the non-physical negative time root:

$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

• Worked cliff problem ($+35.0\,\text{m/s}$ launch from a $200\,\text{m}$ cliff with $g = -9.8\,\text{m/s}^2$):

$-200 = 35t - 4.9t^2 \implies 4.9t^2 - 35t - 200 = 0$

$t = \frac{35 \pm \sqrt{(-35)^2 - 4(4.9)(-200)}}{2(4.9)} = \frac{35 + \sqrt{1225 + 3920}}{9.8} = \frac{35 + 71.7}{9.8} = 10.9\,\text{s}$

• Additional comparison: A purely dropped object from the same $200\,\text{m}$ cliff takes $t = \sqrt{\frac{2(200)}{9.8}} = 6.4\,\text{s}$. To land simultaneously with the upward-thrown object, it must be released $10.9 - 6.4 = 4.5\,\text{s}$ later.

• For a launch of $+25.0\,\text{m/s}$ from an $80\,\text{m}$ cliff, the flight time is $7.3\,\text{s}$, and a dropped object takes $4.0\,\text{s}$ (requiring release $3.3\,\text{s}$ later).

Projectile Motion: Horizontal Launches

• Independence of Orthogonal Kinematic Dimensions:

  • Projectile motion combines uniform horizontal velocity with uniform vertical gravitational acceleration.

  • Horizontal component ($x$): Neglecting air drag, $a_x = 0$. Velocity remains invariant throughout flight:

$v_x = v_{ix} = \text{constant}$

$x = v_{ix} t$

• Vertical component ($y$): Free fall under gravity with zero initial vertical velocity ($v_{iy} = 0$):

$a_y = g$

$y = \frac{1}{2}gt^2$

• Temporal coupling: The flight duration $t$ connects the two independent dimensions:

$t = \sqrt{\frac{2y}{g}}$

• Resultant impact velocity vector:

$v_{fx} = v_{ix}$

$v_{fy} = gt = \sqrt{2gy}$

$v_f = \sqrt{v_{fx}^2 + v_{fy}^2}$

• Critical Conceptual Deductions for Horizontal Launches:

  • An object dropped from rest and an object launched with high horizontal speed from the identical height impact level ground at the exact same instant.

  • Forward launch speed affects range ($x$), but does not alter the vertical time of fall ($t$).

  • The horizontally launched projectile possesses greater total speed at impact because it retains its horizontal velocity component in addition to identical vertical velocity.

  • Nonlinear dependence on elevation: Range scales proportionally with the square root of launch elevation ($x \propto \sqrt{y}$). Doubling height increases range and time by a factor of $\sqrt{2} \approx 1.414$, not by a factor of 2.

• Worked Horizontal Launch Examples:

  • Ball Rolling Off Table ($g = 9.8\,\text{m/s}^2$): Elevation $y = 0.44\,\text{m}$, launch velocity $v_{ix} = 0.24\,\text{m/s}$:

$t = \sqrt{\frac{2(0.44)}{9.8}} = 0.30\,\text{s}$

$x = 0.24(0.30) = 0.072\,\text{m}$

• Cheetah Leaping From Cliff ($v_{ix} = 15.0\,\text{m/s}$, $g = 9.8\,\text{m/s}^2$):

  • Cliff depth of $45\,\text{m}$:

$t = \sqrt{\frac{2(45)}{9.8}} = 3.03\,\text{s}$

$x = 15.0(3.03) = 45.5\,\text{m}$

• Cliff depth doubled to $90\,\text{m}$:

$t = \sqrt{\frac{2(90)}{9.8}} = 4.29\,\text{s}$

$x = 15.0(4.29) = 64.4\,\text{m}\quad (\text{not } 91\,\text{m},\text{ because } x \propto \sqrt{y})$

• Landing Speed Calculations Without Prior Time Resolution ($g = 9.8\,\text{m/s}^2$):

  • Ring projected at $13.0\,\text{m/s}$ from a $200\,\text{m}$ tower:

$v_{fy} = \sqrt{2(9.8)(200)} = 62.6\,\text{m/s}$

$v_f = \sqrt{(13.0)^2 + (62.6)^2} \approx 64\,\text{m/s}$

• Ring projected at $42.0\,\text{m/s}$ from a $150\,\text{m}$ tower:

$v_{fy} = \sqrt{2(9.8)(150)} = 54.2\,\text{m/s}$

$v_f = \sqrt{(42.0)^2 + (54.2)^2} \approx 69\,\text{m/s}$

• Standardized Examination Parameters ($g = 10\,\text{m/s}^2$): Projectile launched at $20\,\text{m/s}$ off an $80\,\text{m}$ cliff:

$t = \sqrt{\frac{2(80)}{10}} = 4\,\text{s}$

$x = 20(4) = 80\,\text{m}$

$v_{fy} = 10(4) = 40\,\text{m/s}$

$v_f = \sqrt{(20)^2 + (40)^2} = \sqrt{400 + 1600} = \sqrt{2000} = 44.7\,\text{m/s}$

Projectile Motion: Launches at an Angle and the Range Equation

• Component Resolution of Launch Velocity:

  • For launch angle $\theta$ above the horizontal:

$v_{ix} = v_i \cos(\theta)$

$v_{iy} = v_i \sin(\theta)$

• Horizontal dynamics ($a_x = 0$):

$x = v_{ix} t$

• Vertical dynamics ($a_y = -g$):

$t_{\text{up}} = \frac{v_{iy}}{g}$

$t_{\text{total}} = \frac{2v_{iy}}{g}$

$h_{\max} = \frac{v_{iy}^2}{2g}$

• Dynamic parameters at the apex of flight:

  • Instantaneous vertical velocity is zero ($v_y = 0$).

  • Instantaneous horizontal velocity remains unchanged ($v_x = v_{ix} \ne 0$).

• Resultant speed at peak equals the horizontal component ($v = v_{ix}$).

• Acceleration at apex equals downward gravitational acceleration ($a = g\,\text{downward}$).

• Worked Example for an Angled Launch ($g = 10\,\text{m/s}^2$):

  • Projectile fired at $50\,\text{m/s}$ at an angle of $30^\circ$ over level terrain:

$v_{ix} = 50 \cos(30^\circ) = 43.3\,\text{m/s}$

$v_{iy} = 50 \sin(30^\circ) = 25\,\text{m/s}$

$t_{\text{total}} = \frac{2(25)}{10} = 5\,\text{s}$

$h_{\max} = \frac{(25)^2}{2(10)} = \frac{625}{20} = 31.25\,\text{m}$

$x = 43.3(5) = 216.5\,\text{m}$

• The Range Equation:

  • Strict Boundary Condition: Valid exclusively over level ground where launch and landing elevations are equal ($\Delta y = 0$). It is invalid for cliff launches or uneven terrain.

  • Analytic derivation:

$0 = v_{iy} t - \frac{1}{2}gt^2 \implies t = \frac{2v_{iy}}{g}$

$x = v_{ix} t = (v_i \cos\theta)\left(\frac{2v_i \sin\theta}{g}\right) = \frac{v_i^2}{g}(2\sin\theta\cos\theta)$

• Applying the trigonometric identity $2\sin\theta\cos\theta = \sin(2\theta)$, the formula simplifies to:

$x = \frac{v_i^2}{g}\sin(2\theta)$

• Consequences of the Range Equation:

  • Maximum Range Condition: The term $\sin(2\theta)$ reaches its maximum value of $1$ when $2\theta = 90^\circ$, proving that $\theta = 45^\circ$ maximizes horizontal range:

$x_{\max} = \frac{v_i^2}{g}$

• Complementary Launch Angles: Angles that sum to $90^\circ$ ($\theta$ and $90^\circ - \theta$, such as $30^\circ$ and $60^\circ$, or $15^\circ$ and $75^\circ$) yield identical horizontal ranges, because $\sin(2\theta) = \sin(180^\circ - 2\theta)$.

• The steeper angle stays in flight longer and reaches a greater maximum altitude.

• Quadratic velocity scaling: Range scales with the square of the launch velocity ($x \propto v_i^2$). Doubling launch speed produces a four-fold ($4\times$) increase in range.

  • Numerical Evaluation of Range at Various Angles ($v_i = 20\,\text{m/s}$, $g = 10\,\text{m/s}^2$):

• Maximum range at $45^\circ$:

$x = \frac{(20)^2}{10}\sin(90^\circ) = 40(1) = 40\,\text{m}$

• Complementary pair $30^\circ$ and $60^\circ$:

$x = 40\sin(60^\circ) = 40\sin(120^\circ) = 40(0.866) = 34.6\,\text{m}$

• Complementary pair $15^\circ$ and $75^\circ$:

$x = 40\sin(30^\circ) = 40\sin(150^\circ) = 40(0.5) = 20\,\text{m}$

• Systematic Comparison of Launch Angles (Identical Launch Speed):

  • Launch angle of $30^\circ$: Range matches $60^\circ$; Max height is lowest; Hang-time is shortest; $v_{ix}$ is largest; $v_{iy}$ is smallest.

  • Launch angle of $45^\circ$: Range is maximized; Max height is intermediate; Hang-time is intermediate; Components are equal ($v_{ix} = v_{iy}$).

  • Launch angle of $60^\circ$: Range matches $30^\circ$; Max height is greatest; Hang-time is longest; $v_{ix}$ is smallest; $v_{iy}$ is largest.

  • Physical explanation: Steeper launch vectors partition greater magnitude into vertical velocity ($v_{iy}$), prolonging hang-time and altitude at the expense of forward speed ($v_{ix}$).

• Core Conceptual Reasoning Rules:

  • Rule 1 (Height dictates time): Matching maximum altitudes imply matching flight times and matching initial vertical velocities ($v_{iy}$), since $h = \frac{v_{iy}^2}{2g}$ and $t = \frac{2v_{iy}}{g}$.

  • Rule 2 (Vertical component drives vertical motion): Greater $v_{iy}$ guarantees both greater peak height and greater flight time.

  • Rule 3 (Horizontal component drives range): For a fixed time of flight, horizontal range is directly proportional to horizontal speed ($x = v_{ix}t$).

  • Rule 4 (Unequal ranges with equal times): If projectiles share identical maximum heights but achieve different ranges, their vertical velocities are identical while their horizontal velocities differ. The projectile with the greatest range has the largest $v_{ix}$ and therefore the largest overall launch speed ($v_i = \sqrt{v_{ix}^2 + v_{iy}^2}$).

  • Rule 5 (Rankings for equal-height trajectories): For multiple projectiles $A$, $B$, and $C$ reaching identical apex heights but landing at distances $A < B < C$:

  • Flight times: $t_A = t_B = t_C$

  • Initial vertical velocities: $v_{iy,A} = v_{iy,B} = v_{iy,C}$

  • Initial horizontal velocities: $v_{ix,A} < v_{ix,B} < v_{ix,C}$

  • Overall initial launch speeds: $v_{i,A} < v_{i,B} < v_{i,C}$

  • Ranking arrows launched at $60^\circ$, $45^\circ$, and $30^\circ$ with equal speeds:

  • Longest time in flight: $60^\circ$

  • Greatest altitude achieved: $60^\circ$

  • Greatest horizontal range: $45^\circ$

  • Greatest horizontal velocity component ($v_{ix}$): $30^\circ$

Comprehensive Formula Reference Sheet

• Measurement and Conversions:

  • Velocity: $\text{m/s}$

  • Acceleration: $\text{m/s}^2$

  • Conversion: $\text{km/h} \div 3.6 = \text{m/s}$

  • Metric prefix milli: $10^{-3}$

• Vector Algebra:

  • Cartesian components: $A_x = A \cos(\theta)$, $A_y = A \sin(\theta)$

  • Resultant magnitude: $R = \sqrt{R_x^2 + R_y^2}$

  • Angle of resultant: $\theta = \tan^{-1}\left(\frac{R_y}{R_x}\right)$

  • Range of two concurrent vectors: $|A - B| \le R \le A + B$

  • Static equilibrium condition: $\vec{R} = 0$

• Motion Graphs:

  • Slope of $x\text{--}t$ equals velocity ($v$

  • Slope of $v\text{--}t$ equals acceleration ($a$

  • Area under $v\text{--}t$ equals displacement ($\Delta x$

  • Area under $a\text{--}t$ equals change in velocity ($\Delta v$

  • Same algebraic signs ($v$ and $a$): speeding up; Opposite signs: slowing down.

• One-Dimensional Kinematics (Uniform Acceleration):

$v_f = v_i + at$

$d = v_i t + \frac{1}{2}at^2$

$v_f^2 = v_i^2 + 2ad$

$d = \left(\frac{v_i + v_f}{2}\right)t$

• Free Fall ($g = 10\,\text{m/s}^2$):

  • Released from rest ($v_{iy} = 0$):

$y = \frac{1}{2}gt^2$

$t = \sqrt{\frac{2y}{g}}$

$v = gt$

$v = \sqrt{2gy}$

• Vertical launches straight up:

$t_{\text{total}} = 2 t_{\text{up}}$

$v_{\text{top}} = 0\,\text{m/s}$

$a_{\text{top}} = g\,\text{downward}$

$h = \frac{v_i^2}{2g}$

• Horizontal Projectile Launches:

$v_{iy} = 0$

$t = \sqrt{\frac{2y}{g}}$

$x = v_{ix} t$

$v_f = \sqrt{v_{ix}^2 + (gt)^2}$

• Angled Projectile Launches:

$v_{ix} = v_i \cos(\theta)$

$v_{iy} = v_i \sin(\theta)$

$t = \frac{2v_{iy}}{g}$

$h = \frac{v_{iy}^2}{2g}$

$x = v_{ix} t$

• The Range Equation (Level Ground Only):

$x = \frac{v_i^2}{g}\sin(2\theta)$

$x_{\max} = \frac{v_i^2}{g}\quad (\text{at } \theta = 45^\circ)$

• Quadratic Formula for Elevated Launches:

$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Practice Examination and Fully Worked Solutions

• Problem 1: Convert a speed of $90\,\text{km/h}$ into $\text{m/s}$.

  • Solution: Divide by the conversion constant $3.6$:

$90 \div 3.6 = 25\,\text{m/s}$

• Problem 2: Two concurrent forces measure $6\,\text{N}$ and $9\,\text{N}$. Determine the full range of physically possible resultants.

  • Solution:

$R_{\min} = 9 - 6 = 3\,\text{N}$