Comprehensive Study Notes on Two-Dimensional Projectile Motion and Kinematics

Course Announcements & Schedule

• Test 1 Schedule:
  • Test 1 will take place next Wednesday (a week from tomorrow, not tomorrow).
  • There will be no live lab session on that day; the only scheduled activity is Test 1.
• Upcoming Quiz:
  • A quiz will be conducted in the next class on Thursday.
• Homework Package:
  • The main homework package will be due next Saturday (not this Saturday).

Fundamentals of Two-Dimensional Projectile Motion

• Independence of Kinematic Axes:
  • In two-dimensional ($2D$) motion, the horizontal ($x$) and vertical ($y$) axes can be analyzed completely independently.
  • The $x$ axis features constant velocity motion because there is no acceleration in the horizontal direction ($a_x = 0$).
  • The $y$ axis functions as standard 1D free-fall motion with vertical acceleration equal to gravity ($a_y = -g = -9.8\,m/s^2$).
• Trajectory Peak Analysis:
  • The maximum height or peak of a trajectory is the optimal starting point for analyzing projectile questions.
  • In 1D vertical free fall, the velocity at maximum height is strictly $0\,m/s$.
  • In $2D$ motion, every point on the trajectory possesses two coordinates $(x, y)$ and two velocity components $(v_x, v_y)$.
  • At maximum height in $2D$ motion, the vertical velocity component is zero ($v_y = 0\,m/s$), but the horizontal velocity component ($v_x$) remains non-zero and unchanged.
• Variables in Two-Dimensional Trajectories:
  • Time ($t$
  • Maximum height ($h$
  • Total time of flight ($T$
  • Horizontal range ($R$
• Complementary Launch Angles and Horizontal Range:
  • If the initial launch speed ($v$) is kept constant, launching a projectile at complementary angles (angles whose sum equals $90^\circ$) produces identical horizontal ranges ($R$).
  • Examples of complementary angle pairs yielding the same horizontal range:
    • $30^\circ$ and $60^\circ$
    • $15^\circ$ and $75^\circ$
    • $40^\circ$ and $50^\circ$

Kinematic Motion Analysis at Time Interval t = 0.12 s

• Problem Statement:
  • An object is thrown upward at an angle of $30^\circ$ above the horizontal with an initial velocity magnitude of $12\,m/s$. It travels along its trajectory to reach point C at time $t = 0.12\,s$. Determine the horizontal velocity, vertical velocity, and overall velocity state at point C.
• Initial Component Calculations:
  • Horizontal velocity component:         $v_x = v \cos(\theta) = 12 \cos(30^\circ) = 10.39\,m/s$
  • Initial vertical velocity component:         $v_{y,i} = v \sin(\theta) = 12 \sin(30^\circ) = 6\,m/s$
• Velocity Evaluation at $t = 0.12\,s$:
  • Horizontal velocity component ($v_x$) remains constant throughout projectile flight:         $v_x = 10.39\,m/s$
  • Final vertical velocity component ($v_{y,f}$) at $t = 0.12\,s$ is calculated using 1D kinematics:         $v_{y,f} = v_{y,i} - g t$$v_{y,f} = 6 - (9.8 \times 0.12) = 6 - 1.176 = 4.82\,m/s$
• Key Results:
  • Horizontal velocity at point C: $10.39\,m/s$
  • Vertical velocity at point C: $4.82\,m/s$
  • Calculated magnitude angle context: $11.5$, corresponding to an orientation angle of approximately $25^\circ$

Initial Velocity Analysis of Grasshopper Jump

• Problem Statement:
  • A grasshopper jumps from the ground at an angle of $42^\circ$ with respect to the horizontal and reaches a maximum vertical height of $0.86\,m$. Calculate the initial velocity $v$ with which the grasshopper left the ground.
• Vertical Axis Analysis:
  • Isolate the vertical motion ($y$ axis) as a 1D free-fall problem.
  • At maximum height ($h_{max} = 0.86\,m$), the final vertical velocity is $v_{y,f} = 0\,m/s$
• Kinematic Formulation:
  • Using the vertical kinematic equation:         $v_{y,f}^2 = v_{y,i}^2 - 2 g h_{max}$
  • Substituting vertical initial velocity $v_{y,i} = v \sin(42^\circ)$:\n        0 = (v \sin(42^\circ))^2 - 2(9.8)(0.86)
  • Solving this single-variable relation yields the initial launch velocity $v$

Inclined Plane Trajectory and Distance Calculation

• Problem Statement:
  • An object jumps horizontally from the top of an inclined ski slope with a surface angle of $35^\circ$ below the horizontal and lands at point Q further down the slope. Determine the total distance $d$ along the incline from the launch point to landing point Q.
• Geometric Coordinate Relationships:
  • Define the coordinate system relative to the origin at the top launch point.
  • Horizontal displacement along the incline:         $x = d \cos(35^\circ)$
  • Vertical drop along the incline:         $y = d \sin(35^\circ)$
• Kinematic Equations:
  • Horizontal constant-velocity motion:         $x = v_x t \implies t = \frac{x}{v_x} = \frac{d \cos(\theta)}{v_x}$
  • Vertical free-fall displacement:         $y = \frac{1}{2} g t^2$
• Algebraic Derivation for Distance $d$:
  • Substitute $y = d \sin(\theta)$ and $t = \frac{d \cos(\theta)}{v_x}$ into the vertical displacement equation:         $d \sin(\theta) = \frac{1}{2} g \left(\frac{d \cos(\theta)}{v_x}\right)^2$
  • Expand the squared term:         $d \sin(\theta) = \frac{1}{2} g \frac{d^2 \cos^2(\theta)}{v_x^2}$
  • Divide both sides by $d$ to eliminate quadratic power:         $\sin(\theta) = \frac{1}{2} g \frac{d \cos^2(\theta)}{v_x^2}$
  • Rearrange to solve explicitly for distance $d$:         $d = \frac{2 v_x^2 \sin(\theta)}{g \cos^2(\theta)}$

Fire Hose Trajectory and Vertical Wall Impact Height

• Problem Statement:
  • A firefighter directs a hose at an angle of $30^\circ$ relative to the ground towards a house located at a horizontal distance of $50\,m$ away. Water emerges from the hose at a velocity of $40\,m/s$ and strikes point P on the wall of the building. Determine the vertical height $y$ at which the water hits the wall. (Note: Point P is not the maximum height of the trajectory).
• Step-by-Step Mathematical Solution:
  • Step 1: Calculate initial horizontal and vertical velocity components:         $v_x = v \cos(30^\circ) = 40 \cos(30^\circ) = 34.64\,m/s$$v_{y,i} = v \sin(30^\circ) = 40 \sin(30^\circ) = 20\,m/s$
  • Step 2: Calculate time $t$ required for the water to traverse the horizontal distance $x = 50\,m$:         $x = v_x t$$50 = (40 \cos(30^\circ)) t$$t = \frac{50}{40 \cos(30^\circ)} = 1.44\,s$
  • Step 3: Calculate vertical position $y$ at $t = 1.44\,s$:         $y = v_{y,i} t - \frac{1}{2} g t^2$$y = (20)(1.44) - 4.9 (1.44)^2$$y = 28.8 - 4.9 (2.0736) = 28.8 - 10.16 = 18.64\,m$
• Final Answer:
  • The water hits the vertical wall of the building at a height of $18.64\,m$

Range Equation and Total Time of Flight Derivation

• Problem Statement:
  • An object is projected at an launch angle of $30^\circ$ with initial velocity $v$ such that it achieves a horizontal distance (range) of $22\,m$. Find the initial velocity $v$, time to reach peak height $t$, and total time of flight $T$
• Velocity Determination via Range Equation:
  • Standard horizontal range equation:         $R = \frac{v^2 \sin(2\theta)}{g}$
  • Substitute given parameters ($R = 22\,m$, $\theta = 30^\circ$, $2\theta = 60^\circ$, $g = 9.8\,m/s^2$):         $22 = \frac{v^2 \sin(60^\circ)}{9.8}$
  • Rearrange and compute initial launch speed $v$:         $v^2 = \frac{22 \times 9.8}{\sin(60^\circ)}$$v = \sqrt{\frac{215.6}{0.866025}} = 15.78\,m/s$
• Time of Peak and Time of Flight Computations:
  • Time to reach maximum height ($t_{peak}$):         $t_{peak} = \frac{v \sin(30^\circ)}{g} = \frac{15.78 \sin(30^\circ)}{9.8}$
  • Total time of flight ($T$) obtained by doubling peak time:         $T = 2 \times t_{peak} = 2 \times \left(\frac{15.78 \sin(30^\circ)}{9.8}\right)$