AP Physics

One-Dimensional Free Fall Motion

• Gravitational Acceleration Dynamics:

  • Near the surface of the Earth, an object in free fall experiences constant downward acceleration due to gravity, represented by $g = -10\,\text{m/s}^2$ (or $-10\,\text{m/s/s}$).

  • Neglecting air resistance, every falling object accelerates downward at this constant rate regardless of its mass.

• Fundamental Kinematic Equations for Dropped Objects:

  • For an object dropped from rest ($v_0 = 0\,\text{m/s}$) at an initial height designated as $x_0 = 0\,\text{m}$:

  • Change in velocity: $\Delta v = a \cdot t = 10\,\text{m/s}^2 \cdot 1\,\text{s}$

  • Position via average velocity: $x = v_{\text{avg}} \cdot t$

  • Position via acceleration and time: $x = \frac{1}{2} a t^2$

  • Magnitude of fall displacement ($x_{\text{fall}}$): $x_{\text{fall}} = 5 t^2$

• Comprehensive Free Fall Motion Table ($0\,\text{s}$ to $5\,\text{s}$):

  • At $t = 0\,\text{s}$: Velocity $v = 0\,\text{m/s}$, Falling Position $x = 0\,\text{m}$.

  • At $t = 1\,\text{s}$: Velocity $v = -10\,\text{m/s}$, Falling Position $x = -5\,\text{m}$ ($x_{\text{fall}} = 5(1)^2 = 5\,\text{m}$).

  • At $t = 2\,\text{s}$: Velocity $v = -20\,\text{m/s}$, Falling Position $x = -20\,\text{m}$ ($x_{\text{fall}} = 5(2)^2 = 20\,\text{m}$).

  • At $t = 3\,\text{s}$: Velocity $v = -30\,\text{m/s}$, Falling Position $x = -45\,\text{m}$ ($x_{\text{fall}} = 5(3)^2 = 45\,\text{m}$).

  • At $t = 4\,\text{s}$: Velocity $v = -40\,\text{m/s}$, Falling Position $x = -80\,\text{m}$ ($x_{\text{fall}} = 5(4)^2 = 80\,\text{m}$).

  • At $t = 5\,\text{s}$: Velocity $v = -50\,\text{m/s}$, Falling Position $x = -125\,\text{m}$ ($x_{\text{fall}} = 5(5)^2 = 125\,\text{m}$).

• Graphical Representation of Free Fall:

  • Velocity vs. Time Graph: A straight line with a constant negative slope. The slope represents the object's acceleration: $\text{slope} = a = -10\,\text{m/s}^2$.

  • Position vs. Time Graph: A downward-opening parabolic curve starting at $(0\,\text{s}, 0\,\text{m})$. The instantaneous slope at any point along the curve represents the object's instantaneous velocity: $\text{slope} = v$.

Two-Dimensional Projectile Motion Fundamentals

• Independence of Horizontal and Vertical Motion:

  • Two-dimensional projectile motion is decomposed into two completely independent one-dimensional motions: horizontal ($x$) and vertical ($y$).

  • Horizontal Acceleration: $a_x = 0\,\text{m/s}^2$. Horizontal velocity ($v_x$) remains constant throughout the entire flight.

  • Vertical Acceleration: $a_y = -g = -10\,\text{m/s}^2$. Vertical velocity ($v_y$) changes continuously due to gravity.

• Key Analytical Formulas:

  • Time to Peak Height: $t_{\text{top}} = \frac{v_{y0}}{g}$

  • Total Time of Flight (Level Ground): $t_{\text{total}} = \frac{2 v_{y0}}{g}$

  • Horizontal Range / Displacement: $X = v_x \cdot t_{\text{air}}$

• Vector Characteristics Along the Parabolic Trajectory:

  • At all points along the path, the horizontal velocity vector $v_x$ maintains a uniform length and direction (pointing horizontally rightward).

  • At the apex (maximum height), the vertical velocity component is exactly zero ($v_y = 0\,\text{m/s}$), while horizontal velocity $v_x$ remains active.

  • The gravitational acceleration vector $g$ points vertically downward with constant magnitude at every point during ascent, peak, and descent.

Diagonal Projectile Launch Analysis

• Launch Parameters & Initial Component Resolution:

  • A cannonball launched diagonally from ground level with an initial speed $v_0 = 36\,\text{m/s}$ at launch angle $56.3^\circ$ resolves into components:

  • Initial horizontal velocity: $v_x = 20\,\text{m/s}$

  • Initial vertical velocity: $v_{y0} = 30\,\text{m/s}$

  • Accelerations: $a_x = 0\,\text{m/s}^2$, $a_y = -10\,\text{m/s}^2$

• Time of Flight and Total Horizontal Range:

  • Total time in air: $t_{\text{air}} = 2 \cdot \frac{v_{\text{vert}}}{g} = 2 \cdot \frac{30\,\text{m/s}}{10\,\text{m/s}^2} = 6\,\text{s}$

  • Total horizontal range ($X$): $X = v_H \cdot t_{\text{air}} = 20\,\text{m/s} \cdot 6\,\text{s} = 120\,\text{m}$

• Second-by-Second Kinematic State Table:

  • Rounding speeds and heights to whole numbers, neglecting initial cannon height and drag (entering zero after landing):

  • $t = 0\,\text{s}$: $v_x = 20\,\text{m/s}$, $v_y = 30\,\text{m/s}$, Height $y = 0\,\text{m}$

  • $t = 1\,\text{s}$: $v_x = 20\,\text{m/s}$, $v_y = 20\,\text{m/s}$, Height $y = 25\,\text{m}$

  • $t = 2\,\text{s}$: $v_x = 20\,\text{m/s}$, $v_y = 10\,\text{m/s}$, Height $y = 40\,\text{m}$

  • $t = 3\,\text{s}$: $v_x = 20\,\text{m/s}$, $v_y = 0\,\text{m/s}$, Height $y = 45\,\text{m}$

  • $t = 4\,\text{s}$: $v_x = 20\,\text{m/s}$, $v_y = -10\,\text{m/s}$, Height $y = 40\,\text{m}$

  • $t = 5\,\text{s}$: $v_x = 20\,\text{m/s}$, $v_y = -20\,\text{m/s}$, Height $y = 25\,\text{m}$

  • $t = 6\,\text{s}$: $v_x = 20\,\text{m/s}$ (or $0\,\text{m/s}$ upon landing), $v_y = -30\,\text{m/s}$ (or $0\,\text{m/s}$ upon landing), Height $y = 0\,\text{m}$

  • Direction of motion vector at $t = 1\,\text{s}$: Points diagonally upward at $45^\circ$ with equal horizontal ($20\,\text{m/s}$) and vertical ($20\,\text{m/s}$) components.

Horizontal Cannonball Launch Dynamics & Proportional Scaling

• Horizontal Launch Height Reduction Problem:

  • Problem Setup: A cannonball shot horizontally from height $h$ travels a horizontal distance $x = 6\,\text{m}$. Calculate the new horizontal distance when launched horizontally from height $\frac{1}{3} h$.

  • Mathematical Derivation:

  • Vertical fall time from height $h$: $t_{\text{fall}} = \sqrt{\frac{2h}{g}}$

  • Fall time from height $\frac{1}{3} h$: $t_{\text{fall}}' = \sqrt{\frac{2(\frac{1}{3} h)}{g}} = \sqrt{\frac{1}{3}} \cdot t_{\text{fall}}$

  • New horizontal distance: $x' = v_x \cdot t_{\text{fall}}' = v_x \cdot \sqrt{\frac{1}{3}} t_{\text{fall}} = \sqrt{\frac{1}{3}} \cdot 6\,\text{m} \approx 0.577 \cdot 6\,\text{m} \approx 3.46\,\text{m}$

  • Conclusion: The distance $3.46\,\text{m}$ falls in option C: Between 3m and 6m.

• Horizontal Velocity Doubling Problem:

  • Problem Setup: A cannonball is shot horizontally with velocity $v$, traveling horizontal distance $R$ in time $t$ before hitting the ground. Determine horizontal distance and time of flight if velocity is doubled to $2v$ from the same height.

  • Mathematical Derivation:

  • Time of flight depends strictly on vertical height $h$ ($t = \sqrt{\frac{2h}{g}}$), so time of flight remains $t$

  • New horizontal distance: $R' = (2v) \cdot t = 2(v \cdot t) = 2R$

  • Conclusion: Correct Option A (Horizontal Distance = $2R$, Time of Flight = $t$).

• Velocity Scaling in Angular Launches:

  • When launch velocity $V$ is scaled by a factor of $\frac{1}{3}$ at a fixed launch angle:

  • Time of flight scales linearly: $t_2 = \frac{1}{3} t_1$

  • Range scales quadratically: $X_2 = \left(\frac{1}{3}\right)^2 X_1 = \frac{1}{9} X_1$

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