Projectile Motion and Kinematics

Historical Perspectives and Theoretical Development of Motion

• Medieval Misconceptions of Projectile Motion:

  • In the Middle Ages, practical applications of projectile warfare—such as launching stones, throwing rocks, and shooting arrows—were well understood, but the scientific explanations behind them were deeply flawed.
  • Drawings and manuscripts dating back to the 1300s illustrate early attempts to explain projectile paths prior to modern observational physics.

• The Impetus Theory:

  • Core Concept: Early scholars believed that air was not empty space, but rather filled with air molecules that actively kept projectiles moving.
  • Proposed Mechanism: As an arrow or stone flew through the air at an inclination angle relative to the horizontal, it was thought to displace air molecules in front of it. These displaced molecules were believed to travel behind the projectile's tail and continuously push it forward.
  • Exhaustion of Impetus: According to impetus theory, when the projectile reached its apex, the air molecules stopped pushing its tail. The impetus was said to be "exhausted," causing the object to lose forward motion and fall straight downward to the ground.

• Deficiencies of Early Theories:

  • Historical explanations lacked observational evidence because tech and tools were unavailable to throw objects to sufficient heights for proper empirical tracking.
  • Scholars lacked fundamental theoretical frameworks, such as Newton's Laws of Motion, which define modern mechanics.

Fundamental Principles and Equations of Projectile Motion

• Parabolic Trajectory:

  • Under constant gravitational acceleration, any ideal projectile follows a parabolic path described mathematically by a downward-facing quadratic function.

• Vector Decomposition of Launch Velocity:

  • A projectile launched with initial velocity $u$ (or $v_0$) at an angle $\theta$ with respect to the horizontal can be broken down into two independent perpendicular velocity components:
  • Horizontal Component ($u_x$ or $v_{0x}$):       u_x = u \n\ncos(\theta)
  • Vertical Component ($u_y$ or $v_{0y}$):       $u_y = u \sin(\theta)$

• Independence of Horizontal and Vertical Motion:

  • Horizontal Direction ($x$-axis):
  • In the absence of air resistance, there is zero horizontal acceleration ($a_x = 0\,m/s^2$).
  • Horizontal velocity remains constant throughout the entire flight: $v_x = v_{0x} = \text{constant}$.
  • Horizontal position equation: $x = v_x t = (v_0 \cos(\theta)) t$.
  • Vertical Direction ($y$-axis):
  • The projectile experiences constant downward acceleration due to Earth's gravity ($a_y = -g$, where $g = 9.81\,m/s^2$ or $g = 9.8\,m/s^2$).
  • As the object ascends, vertical velocity $v_y$ decreases continuously.
  • At the apex/highest point, vertical velocity momentarily drops to zero ($v_y = 0\,m/s$).
  • The projectile does not stop moving entirely at the peak because horizontal velocity $v_x$ remains active.
  • During ascent, vertical motion acts against gravity ($-g$); during descent, vertical motion acts in favor of gravity ($+g$).

• Symmetry of Projectile Motion:

  • When air resistance is neglected, projectile motion is perfectly symmetric in geometry, time, and speed relative to its apex.
  • The time required to ascend to maximum height equals the time required to descend back to launch height.

• Kinematic Equations for Projectile Motion:

  • Horizontal Position:     $x = v_{0x} t$
  • Vertical Position (with positive $y$ defined upward):     $y = y_0 + v_{0y} t - \frac{1}{2} g t^2$
  • Torricelli's Equation for Vertical Motion (time-independent):     $v_y^2 = v_{0y}^2 - 2 g \Delta y$

Derivations of Flight Time, Trajectory, and Range

• Derivation of Time of Flight ($t_{\text{flight}}$):

  • Setting initial elevation $y_0 = 0$ and final elevation $y = 0$ in the vertical position equation:     $0 = v_{0y} t - \frac{1}{2} g t^2$
  • Substituting $v_{0y} = v_0 \sin(\theta)$:     $0 = (v_0 \sin(\theta)) t - \frac{1}{2} g t^2$
  • Factoring out $t$:     $t \left(v_0 \sin(\theta) - \frac{1}{2} g t\right) = 0$
  • Yields two mathematical solutions:
  1. $t = 0$ (instant of launch).
  2. $t_{\text{flight}} = \frac{2 v_0 \sin(\theta_0)}{g}$
  • The factor of $2$ explicitly accounts for the equal duration of upward ascent and downward descent.

• Derivation of Trajectory Equation ($y(x)$):

  • Express time $t$ from the horizontal displacement equation: $t = \frac{x}{v_x} = \frac{x}{v_0 \cos(\theta)}$.
  • Substitute $t$ into the vertical position equation $y = (v_0 \sin(\theta)) t - \frac{1}{2} g t^2$:     $y(x) = (v_0 \sin(\theta)) \left(\frac{x}{v_0 \cos(\theta)}\right) - \frac{1}{2} g \left(\frac{x}{v_0 \cos(\theta)}\right)^2$
  • Simplifying yields the parabolic trajectory equation:     $y(x) = x \tan(\theta) - \frac{g x^2}{2 v_0^2 \cos^2(\theta)}$

• Derivation of Horizontal Range Equation ($R$):

  • Setting $y(x) = 0$ for horizontal distance $x = R$:     $0 = R \tan(\theta) - \frac{g R^2}{2 v_0^2 \cos^2(\theta)}$
  • Solving for non-zero range $R$:     $\tan(\theta) = \frac{g R}{2 v_0^2 \cos^2(\theta)}$$\frac{\sin(\theta)}{\cos(\theta)} = \frac{g R}{2 v_0^2 \cos^2(\theta)}$$R = \frac{2 v_0^2 \sin(\theta) \cos(\theta)}{g}$
  • Using the trigonometric identity $\sin(2\theta) = 2 \sin(\theta) \cos(\theta)$:     $R = \frac{v_0^2 \sin(2\theta)}{g}$

Questions and Discussion

• Distinction Between Trajectory and Range (Dialogue with Lauren):

  • Question: Is trajectory another word for range?
  • Answer: No. Trajectory refers to the entire geometrical path (the curve) traced by the projectile, described by $y$ as a function of $x$ ($y(x)$). Range ($R$) refers strictly to the total horizontal displacement from the launch point to the point where the object returns to baseline elevation ($y = 0$).

• Applicability to Unequal Launch and Landing Heights (Dialogue with Isabelle):

  • Question: Can the standard range formula be used if an object lands at an elevation different from its launch height (e.g., throwing a ball while standing and letting it hit the ground)?
  • Answer: No. The formula $R = \frac{v_0^2 \sin(2\theta)}{g}$ requires equal launch and landing elevations ($y_{\text{initial}} = y_{\text{final}} = 0$). When launch and landing elevations differ, quadratic kinematic position equations must be used instead.

Complete Quantitative Analysis: Projectile Motion Example

• Problem Parameters:

  • Initial launch velocity $v_0 = 15\,m/s$
  • Launch angle $\theta = 37^\circ$

• Part A: Initial Velocity Components and Peak Velocities:

  • Initial Horizontal Component:     $v_{0x} = 15 \times \cos(37^\circ) = 15 \times 0.8 = 12\,m/s$
  • Initial Vertical Component:     $v_{0y} = 15 \times \sin(37^\circ) = 15 \times 0.6 = 9\,m/s$
  • Velocity components at launch ($t = 0\,s$): $v_x = 12\,m/s$, $v_y = 9\,m/s$
  • Velocity components at highest point: $v_y = 0\,m/s$, $v_x = 12\,m/s$

• Part B: Calculating Maximum Height ($\Delta y_{\text{max}}$):

  • Using Torricelli's equation at apex ($v_y = 0\,m/s$):     $v_y^2 = v_{0y}^2 - 2 g \Delta y$$0 = (9)^2 - 2 (9.81) \Delta y$$81 = 19.62 \Delta y \implies \Delta y = \frac{81}{19.62} = 4.13\,m$

• Part C: Calculating Ascent Time and Time of Flight:

  • Time to reach maximum height ($t_{\text{up}}$):     $v_y = v_{0y} - g t$$0 = 9 - 9.81 t \implies t_{\text{up}} = \frac{9}{9.81} = 0.92\,s$
  • Total Time of Flight ($t_{\text{flight}}$):     $t_{\text{flight}} = 2 \times t_{\text{up}} = 2 \times 0.92\,s = 1.84\,s$
  • Direct verification using flight time formula:     $t_{\text{flight}} = \frac{2 \times 15 \times \sin(37^\circ)}{9.81} = 1.84\,s$

• Part D: Calculating Horizontal Range:

  • Method 1 (Horizontal velocity $\times$ flight time):     $R = v_x \times t_{\text{flight}} = 12\,m/s \times 1.84\,s = 22.1\,m$
  • Method 2 (Direct Range equation):     $R = \frac{v_0^2 \sin(2\theta)}{g} = \frac{(15)^2 \sin(74^\circ)}{9.81} = 22.1\,m$

Asymmetric Height Projectile Analysis: Volcano Scenario

• Problem Context:

  • A volcano ejects a rock at initial speed $v_0 = 25\,m/s$ at an angle $\theta = 35^\circ$ above the horizontal.
  • The rock lands at a point $20\,m$ below the initial launch elevation ($y = -20\,m$).

• Step 1: Compute Initial Vertical Velocity Component:   $v_{0y} = 25 \times \sin(35^\circ) = 14.34\,m/s$

• Step 2: Formulate Position Equation:   $y = v_{0y} t - \frac{1}{2} g t^2$$-20 = 14.34 t - 4.90 t^2$$4.90 t^2 - 14.34 t - 20 = 0$

• Step 3: Solve Quadratic Equation for Time ($t$):

  • Solving $4.90 t^2 - 14.34 t - 20 = 0$ yields two roots:     $t \approx 4.0\,s \quad \text{and} \quad t \approx -1.02\,s$
  • The negative root ($t = -1.02\,s$) is discarded as unphysical.
  • The valid duration required for the rock to hit the ground is $t \approx 4.0\,s$

• Step 4: Determine Impact Velocity Components and Direction:

  • Horizontal velocity component (constant throughout flight):     $v_x = 25 \times \cos(35^\circ) = 20.5\,m/s$
  • Vertical velocity component at impact ($t = 4.0\,s$):     $v_y = v_{0y} - g t = 14.34 - (9.81 \times 4.0) = -24.5\,m/s$
  • The negative sign indicates that vertical velocity points downward.
  • The direction angle $\theta$ at impact is negative because it lies below the horizontal axis in the chosen coordinate system.

Optimal Launch Conditions for Maximum Range

• Tactical Archery Context:

  • Launching projectiles (such as arrows) at advancing cavalry requires maximizing horizontal range. Shooting too low causes arrows to fall short; shooting too high allows horses to pass under the flight path.

• Mathematical Proof of Optimal Angle ($\theta = 45^\circ$):

  • Range formula: $R = \frac{v_0^2 \sin(2\theta)}{g}$
  • The function $\sin(2\theta)$ reaches its absolute maximum value of $1$ when:     $2\theta = 90^\circ \implies \theta = 45^\circ$

• Comparison of Launch Angles:

  • Low Angle (e.g., $10^\circ$): Low trajectory, minimal airtime, short horizontal range.
  • High Angle (e.g., $80^\circ$): High vertical peak, long flight time, but low horizontal velocity, resulting in a short horizontal range near the launch point.
  • Optimal Angle ($45^\circ$): Splits launch velocity equally between horizontal ($\cos(45^\circ)$) and vertical ($\sin(45^\circ)$) directions, yielding maximum possible horizontal displacement.

Horizontal Launch Dynamics and Motion Independence

• Horizontal Launch Mechanics:

  • An object launched strictly horizontally off an elevated surface of height $h$ starts with initial horizontal velocity $v_x$ and zero initial vertical velocity ($v_{0y} = 0\,m/s$).
  • The object simultaneously experiences two independent motions: uniform horizontal motion and accelerated vertical free-fall.

• Methods to Increase Range in Horizontal Launches:

  1. Increase initial horizontal velocity ($v_x$).
  2. Increase release height ($h$) of the table/platform.

• Simultaneous Drop vs. Horizontal Launch Demonstration:

  • Experiment: Two identical objects placed at equal height $h$ are released at the exact same instant—one dropped vertically from rest, the other launched horizontally.
  • Outcome: Neglecting air resistance, both objects strike the ground at the exact same instant.
  • Explanation: Downward gravitational acceleration ($g$) acts identically on both objects regardless of horizontal velocity.

• Laboratory Carbon Paper Setup:

  • Metallic balls launched horizontally land on carbon paper overlying blank paper on the floor, leaving precise marks to measure landing positions and calculate horizontal range.

• Horizontal Velocity Non-Zero Proof:

  • Because horizontal acceleration $a_x = 0\,m/s^2$, horizontal velocity $v_x$ remains constant and non-zero throughout flight. Vertical velocity $v_y = 0\,m/s$ occurs exclusively at the apex.

Sample Conceptual and Quantitative Problems

• Problem 1: Speed at Apex for Angled Launch:

  • Question: An object is launched with speed $30\,m/s$ at an angle of $60^\circ$ relative to the horizontal. What is its speed at the highest point?
  • Solution: At peak elevation, $v_y = 0\,m/s$. Total speed equals horizontal velocity $v_x$:     $v = v_x = v_0 \cos(60^\circ) = 30 \times 0.5 = 15\,m/s$

• Problem 2: Launch Angle Determination from Apex Speed:

  • Question: A projectile's launch speed $v_0$ is twice its speed at the highest point. What is its launch angle $\theta$?
  • Solution:
  • Speed at peak: $v_{\text{peak}} = v_x = v_0 \cos(\theta)$.
  • Given condition: $v_0 = 2 v_{\text{peak}} = 2 (v_0 \cos(\theta))$.
  • Divide by $v_0$:       $1 = 2 \cos(\theta) \implies \cos(\theta) = 0.5 \implies \theta = 60^\circ$
  • Example: If launch velocity $v_0 = 20\,m/s$, speed at highest point is $20 \times \cos(60^\circ) = 10\,m/s$.