Exhaustive Study Guide: Two-Dimensional Projectile Motion and Kinematic Proofs

Kinematic Graphs for Asymmetric Projectile Motion

• Physical System Overview

  • A spring-loaded projectile launcher (flip flange) compresses a spring to convert spring potential energy into kinetic energy, launching a projectile that subsequently moves under the influence of gravity alone.
  • In asymmetric projectile motion, the projectile lands at a vertical position lower than its initial launch height ($y_{\text{final}} < y_{\text{initial}}$).

• Horizontal Position vs. Time ($x$ vs. $t$)

  • Because zero horizontal forces act on the projectile (neglecting air resistance), acceleration in the x-direction is zero ($a_x = 0\,m/s^2$).
  • The horizontal velocity ($v_x$) remains strictly constant throughout flight.
  • The $x\text{-position}$ vs. time graph is represented by a straight line with a constant positive slope:     $x(t) = v_{0x} t$

• Vertical Position vs. Time ($y$ vs. $t$)

  • Gravity causes constant downward acceleration ($a_y = -g = -9.8\,m/s^2$).
  • The vertical position graph is an inverted parabola opening downwards.
  • Coordinate System Options:
  • Origin at Launch Point ($y_0 = 0$): The curve starts at $y = 0$, reaches a maximum height ($y_{\text{max}}$), passes through $y = 0$ again, and terminates at a negative position ($y < 0$).
  • Origin at Ground Level ($y_0 > 0$): The curve starts at initial height $y_0$, ascends to $y_{\text{max}}$, and falls until reaching ground level ($y = 0$).

• Horizontal Velocity vs. Time ($v_x$ vs. $t$)

  • $v_x$ is constant and positive throughout the trajectory.
  • The graph is a horizontal straight line situated above the time axis ($v_x > 0$).

• Vertical Velocity vs. Time ($v_y$ vs. $t$)

  • Vertical velocity decreases linearly over time at a constant rate defined by acceleration due to gravity ($a_y = -9.8\,m/s^2$):     $v_y(t) = v_{0y} - g t$
  • The graph is a straight line with a negative slope.
  • Key Graph Features:
  • Starts at a positive initial vertical velocity ($v_{0y} > 0$).
  • Crosses the time axis ($v_y = 0\,m/s$) at peak height.
  • Extends into the negative velocity region as the projectile descends.
  • Because the landing elevation is lower than the launch elevation, the final downward speed ($|v_{y,\text{final}}|$) strictly exceeds the magnitude of the initial vertical speed ($v_{0y}$). Thus, the negative magnitude of $v_y$ at the end of the line is greater than the positive starting value.

Complementary Launch Angles and Dual Solutions

• Dual Launch Angles for Identical Range

  • Firing a projectile at two distinct launch angles with the same initial velocity ($v_0$) can produce the exact same horizontal range ($\Delta x$).
  • For target distance $\Delta x = 1\text{ km} = 1000\text{ m}$ and initial velocity $v_0 = 140\text{ m/s}$, the two valid launch angles are complementary angles:     $\theta_1 = 15^\circ$$\theta_2 = 75^\circ$
  • Complementary angles satisfy the relation:     $\theta_1 + \theta_2 = 90^\circ$

• Mathematical Origin of Dual Solutions

  • Solving for angle $\theta$ yields a trigonometric relationship involving double angles:     $\sin(2\theta) = 0.5$
  • Because the sine function is positive in both the first quadrant ($0^\circ \le \phi \le 90^\circ$) and second quadrant ($90^\circ \le \phi \le 180^\circ$), there are two valid solutions for $2\theta$:     $2\theta_1 = 30^\circ \implies \theta_1 = 15^\circ$$2\theta_2 = 150^\circ \implies \theta_2 = 75^\circ$

• Graphical Representation in Desmos

  • Plotting $y = \sin(2\theta)$ alongside horizontal reference line $y = 0.5$ in degree mode illustrates that the curves intersect twice in the domain $\theta \in [0^\circ, 90^\circ]$:
  • First intersection at $(15^\circ, 0.5)$
  • Second intersection at $(75^\circ, 0.5)$

Flight Time Differences and Target Synchronization

• Time of Flight Equations

  • Flight time as a function of launch angle $\theta$ for level ground is derived by setting vertical displacement to zero:     $t(\theta) = \frac{2 v_0 \sin(\theta)}{g}$
  • Substituting initial conditions ($v_0 = 140\text{ m/s}$, $g = 9.8\text{ m/s}^2$):
  • For $\theta_1 = 15^\circ$:       $t_{15} = \frac{2 \times 140 \times \sin(15^\circ)}{9.8} \approx 7.24\text{ s} \approx 7\text{ s}$
  • For $\theta_2 = 75^\circ$:       $t_{75} = \frac{2 \times 140 \times \sin(75^\circ)}{9.8} \approx 27.58\text{ s} \approx 27\text{ s}$

• Strategy for Simultaneous Impact

  • To strike a target located $1000\text{ m}$ away simultaneously using two shells from the same launcher:
  1. Fire the first shell at the steeper angle ($\theta = 75^\circ$) which remains airborne longer ($t_{75} \approx 27\text{ s}$).
  2. Delay the second launch by the time difference $\Delta t = t_{75} - t_{15} \approx 27\text{ s} - 7\text{ s} = 20\text{ s}$.
  3. Fire the second shell at the shallower angle ($\theta = 15^\circ$).
  4. Both shells hit the target simultaneously at $t = 27\text{ s}$ relative to the initial launch.

• Trajectory vs. Time-Dependent Position Graphs

  • Trajectory Graph ($y$ vs. $x$):
  • A direct photograph/spatial path of motion.
  • $\theta = 15^\circ$ produces a shallow arc with low peak height.
  • $\theta = 75^\circ$ produces a tall, steep arc reaching a much greater maximum height.
  • Both curves begin at $x = 0$ and terminate at the same landing point $x = 1000\text{ m}$.
  • Vertical Position Graph ($y$ vs. $t$):
  • Described by the quadratic function:       $y(t) = v_0 \sin(\theta) t - 4.9 t^2$
  • $\theta = 15^\circ$ curve peaks at approximately $y_{\text{max}} \approx 4\text{ m}$ and returns to zero at $t \approx 7.24\text{ s}$.
  • $\theta = 75^\circ$ curve peaks substantially higher at $y_{\text{max}} \approx 14\text{ m}$ (or higher depending on precise speed) and returns to zero at $t \approx 27.58\text{ s}$.

Non-Quadratic Kinematic Solutions for Target Heights

• Problem Context

  • A football is kicked from an initial height $y_0 = 1\text{ m}$ with launch speed $v_0 = 25\text{ m/s}$ at an angle $\theta = 30^\circ$.
  • Objective: Calculate the speed of the football when it reaches a target height of $y = 5\text{ m}$ on its way up.

• Determining Motion Direction (Upward vs. Downward)

  • A projectile reaches an intermediate altitude $y = 5\text{ m}$ twice (once ascending, once descending).
  • Three Identification Criteria for Upward Path:
  1. Horizontal Displacement ($\Delta x$): Smaller of the two horizontal distances.
  2. Elapsed Time ($t$): Smaller of the two time roots calculated.
  3. Vertical Velocity Component ($v_y$): Vertical velocity is strictly positive ($v_y > 0$).

• Two-Step Method to Avoid Solving Quadratic Equations

  • Direct time-domain solutions require solving $y(t) = y_0 + v_{0y}t - \frac{1}{2}gt^2$ via the quadratic formula, which can be computationally cumbersome.
  • Alternative 2-Step Algorithm:
  • Step 1: Calculate vertical displacement:       $\Delta y = y_{\text{target}} - y_0 = 5\text{ m} - 1\text{ m} = +4\text{ m}$
  • Step 2: Use the time-independent kinematic equation to solve directly for $v_y$:       $v_y^2 = v_{0y}^2 + 2 a_y \Delta y$

• Step-by-Step Calculation

  • Initial vertical velocity component:     $v_{0y} = v_0 \sin(30^\circ) = 25 \times 0.5 = 12.5\text{ m/s}$
  • Initial horizontal velocity component:     $v_x = v_0 \cos(30^\circ) = 25 \times \cos(30^\circ) \approx 21.65\text{ m/s} \approx 21.7\text{ m/s}$
  • Apply kinematic formula for vertical velocity:     $v_y^2 = (12.5)^2 + 2 (-9.8) (+4)$$v_y^2 = 156.25 - 78.4 = 77.85\text{ m}^2/\text{s}^2 \approx 77.8\text{ m}^2/\text{s}^2$
  • Take the square root:     $v_y = \pm \sqrt{77.85} \approx \pm 8.82\text{ m/s}$
  • Select the positive root because the object is ascending:     $v_y = +8.82\text{ m/s}$
  • Combine vector components tip-to-tail to determine total velocity/speed:     $v = \sqrt{v_x^2 + v_y^2} = \sqrt{(21.65)^2 + (8.82)^2} \approx 23.4\text{ m/s}$

Mathematical Proof for Maximum Range Angle ($\theta = 45^\text{o}$)

• Derivation of the Range Equation

  • Over level ground, the horizontal range is given by:     $\Delta x = v_x t = (v_0 \cos\theta) t$
  • The total time of flight $t$ is derived from vertical motion where final height returns to launch height ($\Delta y = 0$):     $v_y = -v_{0y} \implies -v_0 \sin\theta = v_0 \sin\theta - g t$$g t = 2 v_0 \sin\theta \implies t = \frac{2 v_0 \sin\theta}{g}$
  • Substituting $t$ into the range equation:     $\Delta x = (v_0 \cos\theta) \left( \frac{2 v_0 \sin\theta}{g} \right)$$\Delta x = \frac{2 v_0^2}{g} \sin\theta \cos\theta$

• Trigonometric Product Analysis

  • The range expression can be grouped into fixed physical constants and a variable angle function:     $\Delta x = \left( \frac{2 v_0^2}{g} \right) (\sin\theta \cos\theta)$
  • Using double-angle trigonometric identities, $2 \sin\theta \cos\theta = \sin(2\theta)$, which rewrites the range equation as:     $\Delta x = \frac{v_0^2 \sin(2\theta)}{g}$

• Graphical Verification of Maximum

  • Plotting the function $f(\theta) = \sin\theta \cos\theta$:
  • Completes a full cycle in $180^\circ$ (unlike $\sin\theta$ or $\cos\theta$, which require $360^\circ$).
  • For physical projectile launch angles $0^\circ \le \theta \le 90^\circ$, the function achieves its maximum peak at:       $\theta = 45^\circ$
  • Alternatively, analyzing $\sin(2\theta)$, the sine function reaches its absolute maximum value of $1.0$ when its argument equals $90^\circ$:     $2\theta = 90^\circ \implies \theta = 45^\circ$
  • Conclusion: For 2D projectile motion over flat terrain, maximum horizontal range strictly occurs at a launch angle of $\theta = 45^\circ$.