Basic Concepts & Equations of Motion

• Vector perspective:
  • Displacement: $\vec{s} = \vec{v_0}t + \frac{1}{2}\vec{g}t^2$
  • $\vec{g} = (0, -g)$ in standard coordinates ($+y$ upward).
  • Critical reminder: $\vec{g}$ is always downward, so $g$ is positive but acceleration in $y$-direction is negative.
• Component decomposition (standard $x$-$y$ coordinates):
  • Equation:
    • $x(t) = v_0 t \cos\theta$
    • $y(t) = v_0 t\sin\theta - \frac{1}{2} g t^2$
  • Key insights:
    • Horizontal motion ($x$-axis): Constant velocity, no acceleration
    • Vertical motion ($y$-axis): free fall with constant acceleration $-g$.
  • Common confusion:
    • Using $\sin\theta$ for $x$-component or $\cos\theta$ for $y$-component
• Trajectory equation (path shape):
  • Eliminate time $t$ from $x(t)$ and $y(t)$:
    $$ y = x \tan\theta - \frac{g x^2}{2 v_0^2 \cos^2\theta} $$
    • Alternatively, rewrite using $\sec^2\theta = 1 + \tan^2\theta$:
      $$ y = x \tan\theta - \frac{g x^2 (1 + \tan^2\theta)}{2 v_0^2} $$
  • Why this matters: The trajectory is a parabola
• Special cases:
  • Horizontal throw ($\theta = 0^\circ$):
    • $x = v_0 t$, $y = -\frac{1}{2} g t^2$
    • Time to hit ground from height $h$: $t = \sqrt{\frac{2h}{g}}$
    • Critical reminder: Initial vertical velocity is zero, but vertical acceleration is still $-g$.
  • Symmetric diagonal throw (launch/land at same height):
    • Range: $R = \frac{v_0^2}{g} \sin 2\theta\propto \sin 2\theta$
      • Maximum range at $\theta = 45^\circ$ (since $\sin 90^\circ = 1$).
      • Same distance for $\theta = 45^\circ + \alpha$ and $\theta = 45^\circ - \alpha$
    • Common confusion:
      • Using $R = \frac{v_0^2 \sin 2\theta}{g}$ is invalid for asymmetric launches (e.g., off a cliff).
      • Always solve $y(t) = y_{\text{land}}$ for $t$ first.
• Time is universal:
  • $t$ is identical in $x(t)$ and $y(t)$. Never solve for separate times for horizontal/vertical motion.
  • Example: To find range, first solve $y(t) = 0$ for $t$ (using vertical motion), then plug $t$ into $x(t)$.