Exhaustive Study Notes on Projectile Motion and Two-Dimensional Kinematics

Motion Along the X-Axis

• Equation for X Coordinate Change:

  • The equation describing how the x-coordinate changes over time is:
    X = X0 + V{0x} imes t
  • Where:
  • $X_0$ = initial x-coordinate (often taken as 0, if starting from origin)
  • $V_{0x}$ = initial (constant) velocity in the x direction
  • $t$ = time

• Uniform Motion:

  • Motion along the x-axis ($o_x$) is uniform, meaning constant velocity.
  • Example: If $V_{0x} = 25 ext{ m/s}$, it remains constant throughout the motion at $25 ext{ m/s}$.

• Finding Initial Velocity ($V_{0x}$):

  • Given that an object is launched at an angle (e.g., $30^{ ext{o}}$) with an initial speed (e.g., $80 ext{ m/s}$), the component in the x direction is:
    V{0x} = V0 imes ext{cos}( heta) = 80 ext{ m/s} imes ext{cos}(30^{ ext{o}})

• Constant Velocity Explanation:

  • The value of $V_{0x}$ calculated remains constant regardless of the position along the x-axis.

Motion Along the Y-Axis

• Equation for Y Coordinate Change:

  • The equation that describes vertical motion is:
    Y = Y0 + V{0y} imes t - rac{1}{2} g t^2
  • Where (
  • $Y_0$ = initial y-coordinate
  • $V_{0y}$ = initial velocity in the y direction
  • $g$ = acceleration due to gravity (approximately $9.8 ext{ m/s}^2$, taken as negative in this context)
  • The term $- rac{1}{2} g t^2$ represents the displacement due to gravity.

• Accelerated Motion:

  • Unlike motion along the x-axis, motion along the y-axis ($o_y$) is accelerated due to gravitational pull, leading to a change in velocity over time.

• Initial Vertical Velocity ($V_{0y}$):

  • Similar to $V{0x}$, $V{0y}$ is calculated as:
    V{0y} = V0 imes ext{sin}( heta) = 80 ext{ m/s} imes ext{sin}(30^{ ext{o}})
  • Example calculation yields $V_{0y} = 40 ext{ m/s}$.

Equations for Vertical Motion

• Three essential equations address vertical motion:
  1. Velocity Change:
     Vy = V{0y} + g imes t
  2. Energy Conservation:
     Vy^2 = V{0y}^2 + 2g(Y - Y_0)
  3. Position Change (already mentioned):
     Y = Y0 + V{0y} imes t - rac{1}{2} g t^2

Coordinate System

• When defining motion:
  • Initial coordinates ($X0$, $Y0$):
  • Set $X0 = 0$, $Y0 = 0$ for a launch from the origin. In scenarios where a trajectory begins above ground, $Y0$ can be greater than zero (e.g., $Y0 = 5 ext{ m}$ if launched from a height).
• Example with rolling ball: if an object rolls off at $Y0 = 5 ext{ m}$, then $X0 = 0$, $Y_0 = 5$.

Setting Up Problems

• Important Steps:
  1. Draw Motion Diagram
  2. Define Coordinate System (origin, positive x and y directions)
  3. List Known Parameters (initial velocity, angles, heights, etc.) and Unknowns (final velocities, distances, etc.)
  4. Choose Relevant Equations to Apply

Example Application: Vector Calculations

• For an object launched at $30^{ ext{o}}$ with an initial speed of $80 ext{ m/s}$ over $2$ seconds, find:

  • Components of velocity at point P after 2 seconds:
    V{Px} = V{0x}
    V{Py} = V{0y} + g imes t

• For the specified example:

  • Calculate $V{Px}$ = constant, $V{0x} = 69.28 ext{ m/s}$
  • For $V_{Py}$, substitute known values, leading to an answer of $20 ext{ m/s}$ when $t = 2$ seconds.

• Magnitude of Velocity:

  • Use the Pythagorean theorem to find magnitude of velocity:
    V = ext{sqrt}(V{Px}^2 + V{Py}^2)
  • Result yields magnitude (e.g., $72.11 ext{ m/s}$) and direction using inverse tangent:
    hetaP = ext{tan}^{-1} rac{V{Py}}{V_{Px}}

Rolling Ball Example

• Consider a ball rolling off a $5 ext{ m}$ height with a speed of $4 ext{ m/s}$. Define coordinate system.

• Calculate components:

  • $V{0x} = 4 ext{ m/s}$, $V{0y} = 0 ext{ m/s}$

• Find how long it takes to hit the ground by using:
  Y = Y0 + V{0y}t - rac{1}{2}gt^2

  • Solve for $t_P$ from this equation.

• Presuming $Y_0 = 5 ext{ m}$, $Y = 0$, allows simplifying calculations.

  • Find $t = 1 ext{ s}$.

• Calculate distance fallen:
  XP = V{0x} imes t_P

  • Should yield the horizontal distance the ball lands from the edge.