Advanced Topics

• Inclined plane motion:

  • Case 1: Rotated coordinates (simpler for incline problems):
    • Define $x'$ along incline, $y'$ perpendicular to incline.
    • Acceleration components:
      • $a_{x'} = -g \sin\phi$ (parallel to incline)
      • $a_{y'} = -g \cos\phi$ (perpendicular to incline)
    • Why rotate? Simplifies equations for projectiles launched/landing on slopes.
  • **Case 2: Standard coordinates **:
    • Incline equation: $y = x \tan\phi$
    • Solve $y(t) = x(t) \tan\phi$ for $t$, then compute range along incline.
  • Max range:
    • Uphill incline ($\phi > 0$): $\theta = 45^\circ - \frac{\phi}{2}\lt 45^\circ$
    • Downhill ($\phi < 0$): $\theta = 45^\circ + \frac{|\phi|}{2}\gt45^\circ$
    • Use optimal angle to calculate the range

• Envelope curve of all trajectories:

  • For fixed launch speed $v_0$, all possible trajectories are bounded by:
    $$ y = \frac{v_0^2}{2g} - \frac{g x^2}{2 v_0^2} $$
  • Derivation steps:
    1. Start with trajectory equation: $y = x \tan\theta - \frac{g x^2 (1 + \tan^2\theta)}{2 v_0^2}$
    2. Rearrange as quadratic in $\tan\theta$:
       $$ \left( \frac{g x^2}{2 v_0^2} \right) \tan^2\theta - x \tan\theta + \left( y + \frac{g x^2}{2 v_0^2} \right) = 0 $$
    3. For real $\theta$, discriminant $\geq 0$:
       $$ \Delta =(-x)^2 - 4 \left( \frac{g x^2}{2 v_0^2} \right) \left( y + \frac{g x^2}{2 v_0^2} \right) \geq 0 $$
    4. Solve the above: $$ y \le \frac{v_0^2}{2g} - \frac{g x^2}{2 v_0^2} $$
    5. The boundary of reachable points is when discriminant $= 0$ , or :
       $$ y = \frac{v_0^2}{2g} - \frac{g x^2}{2 v_0^2} $$
  • Key insights:
    • Points inside the envelope: reachable with two launch angles.
    • Points on the envelope: reachable with exactly one launch angle (minimum speed required).
    • Points outside: unreachable for given $v_0$.

• Minimal speed to reach a point $(x, y)$:

  • Minimal speed is found by solving discriminant $\Delta = 0$ .
  • Why this works:
    • Discriminant $\Delta$ decreases as $v_0$ decreases. So below critical speed, $\Delta<0$ and no $\theta$ will allow point $(x, y)$ reachable.
    • This speed gives exactly one trajectory (tangent to envelope) that passes through $(x, y)$.
  • It is not solved by $v_y = 0$

• Critical point:

  • Envelope curve is not a trajectory—it’s the boundary of all possible paths for fixed $v_0$.

• Parameterization by launch angle $\theta$ at fixed time $t$:

  • At time $t$, all possible positions for different $\theta$ form a circle:
    $$ x = (v_0 \cos\theta) t, \quad y = (v_0 \sin\theta) t - \frac{1}{2} g t^2 $$
    • Eliminate $\theta$: $x^2 + \left( y + \frac{1}{2} g t^2 \right)^2 = (v_0 t)^2$
  • Physical meaning:
    • Circle centered at $(0, -\frac{1}{2} g t^2)$ with radius $v_0 t$.
    • All projectiles launched at time $t=0$ with speed $v_0$ are on this circle at time $t$.