Unit 2 Two-Dimensional Motion and Projectile Kinematics Study Notes

Independence of Perpendicular Vector Components in Relative Motion

• Core Principle of Perpendicular Vectors:

  • Motion in perpendicular dimensions ($x$ and $y$) operates completely independently.
  • An external influence acting purely along one coordinate axis produces zero component along an orthogonal axis.
  • In relative motion scenarios, such as crossing a river, the velocity of the river current affects downstream displacement but exerts no influence over the velocity component directed perpendicularly across the river.

• Crossing Time Calculation (Boat #1):

  • Given Parameters:
  • River width ($d$): $120\,m$
  • Current velocity ($v_c$): $4\,m/s$
  • Across-river velocity ($v_y$): $3\,m/s$
  • Governing Relationship:$\Delta t = \frac{d}{v_y}$
  • Step-by-Step Solution:$\Delta t = \frac{120\,m}{3\,m/s} = 40\,s$
  • Physical Mechanism: The time required to cross depends exclusively on the distance across and the velocity vector directed across. The downstream current does not assist or impede progress across the river banks.

• Downstream Displacement Calculation (Boat #2):

  • Given Parameters:
  • River width ($d$): $120\,m$
  • Current velocity ($v_c$): $4\,m/s$
  • Across-river velocity ($v_y$): $6\,m/s$
  • Step 1 — Calculate Crossing Time:$\Delta t = \frac{120\,m}{6\,m/s} = 20\,s$
  • Step 2 — Calculate Downstream Drift Distance:$d_x = v_c \times \Delta t$$d_x = (4\,m/s)(20\,s) = 80\,m$
  • The total downstream distance is $80\,m$.

• Proportional Scaling with Current Variations (Boat #3 vs. Boat #2):

  • Given Parameters for Boat #3:
  • River width ($d$): $120\,m$
  • Increased current velocity ($v_{c3}$): $8\,m/s$
  • Across-river velocity ($v_y$): $6\,m/s$
  • Comparative Analysis:
  • Crossing time remains unchanged because the across-river velocity is constant:       $\Delta t = \frac{120\,m}{6\,m/s} = 20\,s$
  • New downstream distance:       $d_{x3} = (8\,m/s)(20\,s) = 160\,m$
  • Ratio between downstream distances:       $\frac{d_{x3}}{d_{x2}} = \frac{160\,m}{80\,m} = 2$
  • Proportionality Rule: When crossing duration is held constant, downstream distance scales linearly with river current velocity ($d_x \propto v_c$). Doubling current speed directly doubles the downstream displacement.

Two-Dimensional Vector Addition and Resultant Displacement

• Vector Decomposition and Component Summation:

  • Two-dimensional displacement requires resolving individual movements into orthogonal Cartesian axes (North-South as the $y$-axis, East-West as the $x$-axis).
  • Opposite directional components are combined algebraically by subtraction; collinear components in the same direction are combined by addition.

• Multi-Leg Walk Path Analysis:

  • Given Movement Segments:
  • $8\,m\text{ North}$
  • $5\,m\text{ East}$
  • $3\,m\text{ South}$
  • $4\,m\text{ East}$
  • Step 1 — Net Vertical ($y$-axis) Displacement:$\Delta y = 8\,m - 3\,m = +5\,m\text{ (North)}$
  • Step 2 — Net Horizontal ($x$-axis) Displacement:$\Delta x = 5\,m + 4\,m = +9\,m\text{ (East)}$
  • Step 3 — Directional Orientation:
  • Because $\Delta y > 0$ (North) and $\Delta x > 0$ (East), the resultant vector is oriented up and to the right, corresponding to the northeast quadrant.
  • Step 4 — Magnitude Calculation via Pythagorean Theorem:
  • Since North and East form a right angle ($90^\circ$), the magnitude ($d$) of total displacement is determined by the hypotenuse:       $d = \sqrt{(\Delta y)^2 + (\Delta x)^2}$$d = \sqrt{(5)^2 + (9)^2} = \sqrt{25 + 81} = \sqrt{106}$$d \approx 10.3\,m$
  • Final Resultant Vector: The student has a total displacement of approximately $10.3\,m\text{ northeast}$.

Fundamental Kinematic Principles of Projectile Motion

• Acceleration Characteristics in Projectile Flight:

  • When aerodynamic drag is negligible, the magnitude and direction of acceleration acting on an airborne object remain constant throughout the entire trajectory.
  • Gravitational acceleration acts continuously downward toward Earth's center:     $a_y = -g \approx -9.8\,m/s^2$
  • At the trajectory apex (highest point), the acceleration does not equal zero. The acceleration magnitude remains strictly $9.8\,m/s^2$ downward.

• Velocity Component Dynamics:

  • Distinction Between Velocity and Acceleration: Acceleration is invariant, whereas the velocity vector alters in both magnitude and direction at every instant.
  • Vertical Component ($v_y$):
  • Ascent Phase: As the projectile rises, gravitational acceleration opposes upward motion, causing the magnitude of $v_y$ to decrease.
  • Trajectory Peak: At the precise apex, vertical velocity vanishes momentarily ($v_y = 0\,m/s$).
  • Descent Phase: As the projectile falls, gravitational acceleration acts in the direction of motion, causing the magnitude of $v_y$ to increase.
  • Overall Trend: The magnitude of the vertical component decreases and then increases over the duration of the trajectory.
  • Horizontal Component ($v_x$):
  • In the absence of horizontal forces ($\Sigma F_x = 0$), horizontal acceleration is zero ($a_x = 0\,m/s^2$).
  • The horizontal component of velocity remains constant throughout the entire duration of the flight.

• Horizontally Launched Projectiles:

  • For an object rolling horizontally off an elevated surface (such as a table at height $h = 2.0\,m$):
  • Horizontal speed remains constant due to the absence of horizontal net force.
  • Vertical speed increases downward from an initial vertical velocity of $v_{iy} = 0\,m/s$ under constant gravitational acceleration ($a_y = -g$).

Scaling Dynamics: Velocity and Height Variations in Horizontal Projectiles

• Governing Kinematic Relations for Horizontal Launches:

  • Vertical displacement from launch height ($h$) with $v_{iy} = 0$:     $h = \frac{1}{2}g t^2$
  • Time in air ($t$):     $t = \sqrt{\frac{2h}{g}}$
  • Horizontal distance / range ($D$):     $D = v_x t$

• Effect of Increasing Horizontal Launch Velocity (Constant Height):

  • Baseline Scenario: A ball rolls off a platform of height $h = 5\,m$ with initial velocity $v_{x1} = 15\,m/s$, covering horizontal distance $D$ in flight time $T$.
  • Modified Condition: Velocity increases to $v_{x2} = 30\,m/s$ while height remains $h = 5\,m$.
  • Flight Time Analysis:$t = \sqrt{\frac{2(5\,m)}{9.8\,m/s^2}} \approx 1.01\,s$
  • Because height is constant, time of flight is unchanged:       $T_{\text{new}} = T$
  • Horizontal Range Analysis:$D_{\text{orig}} = (15\,m/s)(T)$$D_{\text{new}} = (30\,m/s)(T) = 2(15\,m/s)(T)$$D_{\text{new}} = 2D$
  • Conclusion: Doubling initial horizontal velocity doubles horizontal distance ($2D$) while preserving flight time ($T$).

• Effect of Increasing Launch Height (Constant Velocity):

  • Baseline Scenario: A projectile launched from height $h_1 = 5\,m$ with $v_x = 20\,m/s$ has flight time $T$ and range $D$.
  • Modified Condition: Height increases to $h_2 = 20\,m$ while velocity remains $v_x = 20\,m/s$.
  • Height Ratio Factor:$\frac{h_2}{h_1} = \frac{20\,m}{5\,m} = 4$
  • Flight Time Scaling:
  • Flight time scales with the square root of height ($t \propto \sqrt{h}$):       $\frac{T_{\text{new}}}{T} = \sqrt{\frac{h_2}{h_1}} = \sqrt{4} = 2$$T_{\text{new}} = 2T$
  • Horizontal Range Scaling:
  • Because horizontal velocity is unchanged and time doubles:       $D_{\text{new}} = v_x(2T) = 2(v_x T)$$D_{\text{new}} = 2D$
  • Conclusion: Quadrupling the height doubles both the flight duration ($2T$) and the horizontal distance ($2D$).

• Summary Rules of Proportionality:
  • Changing horizontal velocity alone alters horizontal distance linearly while leaving flight time unaffected.
  • Changing launch height alters flight time by the square root of the height factor; horizontal distance scales in direct proportion to the resulting flight time when horizontal speed is constant.

Symmetric Ground-to-Ground Projectile Kinematics

• Temporal Symmetry in Flight:

  • For a projectile launched and landing at identical elevations, the upward flight time equals the downward fall time.
  • Total time of flight ($t_{\text{total}}$) is twice the time required to reach the peak apex ($t_{\text{peak}}$).

• Quantitative Problem Solution ($v_{iy} = 15\,m/s$, $v_x = 8\,m/s$, $g = 10\,m/s^2$):

  • Time in Air Determination:
  • Vertical velocity at maximum height: $v_{fy} = 0\,m/s$.
  • Kinematic equation for vertical ascent:       $v_{fy} = v_{iy} - gt_{\text{peak}}$$0 = 15 - 10t_{\text{peak}}$$10t_{\text{peak}} = 15$$t_{\text{peak}} = \frac{15}{10} = 1.5\,s$
  • Total duration in the air:       $t_{\text{total}} = 2 \times t_{\text{peak}} = 2(1.5\,s) = 3\,s$
  • Horizontal Displacement Determination:
  • Uniform motion along the horizontal axis:       $\Delta x = v_x \times t_{\text{total}}$$\Delta x = (8\,m/s)(3\,s) = 24\,m$

• General Problem-Solving Strategy:

  • Always analyze the vertical motion first to extract the time parameter ($t$).
  • Substitute the resulting time parameter into the horizontal motion equation to determine horizontal range.

Elevated Projectile Launches and Time-Dependent Velocity Vectors

• Velocity Vector State at Arbitrary Flight Times:

  • Launch site: Top of a tall structure.
  • Initial velocity parameters: $v_{0x} = 25\,m/s$, $v_{0y} = 30\,m/s$ upward.
  • Gravitational acceleration: $g = 10\,m/s^2$ downward.
  • Target evaluation time: $t = 2\,s$.

• Component-by-Component Calculation:

  • Horizontal Velocity Component ($v_x$):
  • In the absence of air drag, horizontal acceleration is zero ($a_x = 0$):       $v_x(t) = v_{0x} = 25\,m/s$
  • Vertical Velocity Component ($v_y$):
  • Gravity reduces upward velocity at a rate of $10\,m/s$ per second:       $v_y(t) = v_{0y} - gt$$v_y(2) = 30 - (10)(2)$$v_y(2) = 30 - 20 = 10\,m/s\text{ upward}$
  • Resulting Velocity Components at $t = 2\,s$:
  • Horizontal velocity: $25\,m/s$
  • Vertical velocity: $10\,m/s\text{ upward}$

• Pedagogical Insights:

  • Launch height does not influence velocity components at intermediate times prior to ground contact.
  • Gravitational acceleration influences only vertical velocity components; it has no cross-axis influence on horizontal velocity.

Trajectory Velocity Profiles and Speed Conservation

• Speed and Velocity Differentiation:

  • Velocity ($\mathbf{v}$) is a vector quantity possessing directional components ($v_x$, $v_y$).
  • Speed ($v$) is a scalar quantity defined as the magnitude of the velocity vector:     $v = \sqrt{(v_x)^2 + (v_y)^2}$

• Detailed Flight Profile ($v_0 = 13\,m/s$, $v_x = 5\,m/s$, $v_{0y} = +12\,m/s$):

  • Launch Point:
  • Horizontal velocity: $v_x = 5\,m/s$
  • Vertical velocity: $v_y = +12\,m/s$
  • Total speed:       $v = \sqrt{(5)^2 + (12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13\,m/s$
  • Highest Trajectory Point (Apex):
  • Vertical motion stops momentarily: $v_y = 0\,m/s$.
  • Horizontal motion remains unchanged: $v_x = 5\,m/s$.
  • Total speed at apex:       $v = \sqrt{(5)^2 + (0)^2} = \sqrt{25} = 5\,m/s$
  • Projectile speed reaches a minimum value at the apex, equal entirely to its horizontal velocity.
  • Terminal Flight State (Just Before Impact):
  • Assuming level terrain and negligible air resistance, vertical speed is fully restored in magnitude with reversed sign:       $v_y = -12\,m/s$
  • Horizontal velocity persists:       $v_x = 5\,m/s$
  • Total speed just before impact:       $v = \sqrt{(5)^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13\,m/s$
  • When returned to original launch height, the terminal speed equals the launch speed.

Graphical Motion Analysis and Catapult Trajectory Modeling

• Graphical Integration Foundations:

  • On a velocity-time ($v$-$t$) graph, the slope represents instantaneous acceleration:     $a = \frac{\Delta v}{\Delta t}$
  • The geometric area bounded between the velocity graph curve and the time axis corresponds to displacement ($\Delta x$ or $\Delta y$):     $\text{Displacement} = \int v\,dt$

• Catapult Trajectory 1 Analysis ($v_{0x} = 7\,m/s$, $v_{0y} = 20\,m/s$, $t = 4.0\,s$, $g = 10\,m/s^2$):

  • Horizontal Graph Construction:
  • Horizontal acceleration is zero ($a_x = 0$).
  • Represented by a flat, horizontal line at $v_x = +7\,m/s$ spanning from $t = 0\,s$ to $t = 4.0\,s$.
  • Vertical Graph Construction:
  • Vertical velocity obeys the linear equation:       $v_y(t) = 20 - 10t$
  • Discrete Points:
    • $t = 0\,s \implies v_y = +20\,m/s$
    • $t = 1\,s \implies v_y = +10\,m/s$
    • $t = 2\,s \implies v_y = 0\,m/s$
    • $t = 3\,s \implies v_y = -10\,m/s$
    • $t = 4\,s \implies v_y = -20\,m/s$
  • A straight, downward-sloping line connects $(0, +20)$ to $(4.0, -20)$.
  • The slope is $\frac{-20 - 20}{4.0 - 0} = -10\,m/s^2$, reflecting constant gravitational acceleration.
  • The curve intersects the time axis at $t = 2\,s$, marking the apex where ascent transitions to descent.
  • Maximum Height Calculation via Graph Area:
  • The maximum height corresponds to upward displacement between $t = 0\,s$ and $t = 2\,s$.
  • The region forms a right triangle:       $\text{Base} = 2\,s$$\text{Height} = 20\,m/s$$\text{Area} = \frac{1}{2}(\text{base})(\text{height}) = \frac{1}{2}(2\,s)(20\,m/s) = 20\,m$
  • Maximum height achieved is $20\,m$.
  • Internal Analytical Checks:
  • Kinematic equation verification:       $h = v_{0y}t - \frac{1}{2}gt^2 = (20)(2) - \frac{1}{2}(10)(2)^2 = 40 - 20 = 20\,m$
  • Full-flight vertical displacement verification: The triangular region below the axis from $t = 2\,s$ to $t = 4\,s$ has an area of $\frac{1}{2}(2\,s)(-20\,m/s) = -20\,m$. Total vertical displacement is $+20\,m + (-20\,m) = 0\,m$.
  • Horizontal range verification: The rectangular area under the horizontal velocity graph is $(7\,m/s)(4\,s) = 28\,m$.

• Catapult Trajectory 2 Analysis ($v_{0x} = 5\,m/s$, $v_{0y} = 10\,m/s$, target at $20\,m$):

  • Flight Parameters:
  • Air time: $2.0\,s$
  • Horizontal travel: $10\,m$
  • Velocity Graphs:
  • Horizontal Graph: Flat line at $v_x = +5\,m/s$ across the entire $2.0\,s$ interval ($t = 0.0\,s$, $0.5\,s$, $1.0\,s$, $1.5\,s$, $2.0\,s$ all equal $+5\,m/s$).
  • Vertical Graph: Straight line given by $v_y = 10 - 10t$:
    • $t = 0.0\,s \implies v_y = +10\,m/s$
    • $t = 0.5\,s \implies v_y = +5\,m/s$
    • $t = 1.0\,s \implies v_y = 0\,m/s$
    • $t = 1.5\,s \implies v_y = -5\,m/s$
    • $t = 2.0\,s \implies v_y = -10\,m/s$
  • Maximum Height Calculation via Graph Area:
  • The projectile attains maximum height when vertical velocity reaches zero at $t = 1.0\,s$.
  • Bounded area under curve from $t = 0\,s$ to $t = 1.0\,s$ forms a right triangle:       $\text{Base} = 1.0\,s$$\text{Height} = 10\,m/s$$\text{Area} = \frac{1}{2}(\text{base})(\text{height}) = \frac{1}{2}(1.0\,s)(10\,m/s) = 5\,m$
  • Maximum height is $5\,m$.
  • Target Performance Assessment:
  • Actual projectile range is $x = v_x t = (5\,m/s)(2.0\,s) = 10\,m$.
  • The target is positioned $20\,m$ from the launch site.
  • The catapult launch falls short by $10\,m$ ($20\,m - 10\,m = 10\,m$), failing to strike the intended target.