{
  "title": "Projectile Motion — {{ class_or_section }}",
  "session": "{{ session_id, e.g. MECH-3.2 }}",
  "concepts": ["projectile motion", "kinematics", "independence of perpendicular motion"],
  "slides": [
    {
      "heading": "One throw, two motions",
      "blocks": [
        {
          "md": "Launch a ball at speed $v_0$ and angle $\\theta$. The trick that makes projectiles easy: the **horizontal** and **vertical** motions do not talk to each other. Gravity pulls only downward, so sideways the ball just coasts."
        },
        {
          "md": "Split the launch velocity once, at the start:\n\n$$v_x = v_0 \\cos\\theta, \\qquad v_y = v_0 \\sin\\theta$$",
          "fragment": true
        }
      ],
      "speaker_notes": "Demonstrate with a dropped ball and a thrown ball released at the same instant — they land together. That shared landing time IS the independence claim. Replace {{ class_or_section }} with your period."
    },
    {
      "heading": "The two equations of the flight",
      "blocks": [
        {
          "md": "Horizontal is constant-velocity; vertical is constant-acceleration under $g$:"
        },
        {
          "md": "$$x(t) = v_0 \\cos\\theta \\cdot t, \\qquad y(t) = v_0 \\sin\\theta \\cdot t - \\tfrac{1}{2} g t^2$$"
        },
        {
          "md": "Everything else — how high, how far, how long — is squeezed out of these two by choosing the right moment $t$.",
          "fragment": true
        }
      ],
      "speaker_notes": "Stress that $t$ is the SAME variable in both equations: that is the bridge between the two independent motions. Use $g \\approx 9.8\\ \\mathrm{m\\,s^{-2}}$."
    },
    {
      "heading": "Worked example: the range",
      "blocks": [
        {
          "md": "A ball leaves the ground at $v_0 = {{ launch_speed, e.g. 20 }}\\ \\mathrm{m\\,s^{-1}}$ and $\\theta = {{ angle, e.g. 30 }}^{\\circ}$. How far away does it land?"
        },
        {
          "md": "It lands when $y = 0$ again, at $t = \\dfrac{2 v_0 \\sin\\theta}{g}$. Substitute into $x(t)$ and use $2\\sin\\theta\\cos\\theta = \\sin(2\\theta)$:\n\n$$R = \\frac{v_0^2 \\sin(2\\theta)}{g}$$\n\nPlug in your launch numbers and evaluate. The range is largest at $\\theta = 45^{\\circ}$, where $\\sin(2\\theta) = 1$.",
          "solution": true
        }
      ],
      "speaker_notes": "Let the class derive the landing time before you reveal the solution block. The double-angle step is the one they forget — flag it. Swap the two launch placeholders for concrete values."
    },
    {
      "heading": "Your turn",
      "blocks": [
        {
          "md": "Same launch, new question: what is the **greatest height** the ball reaches?\n\nHint: at the top, $v_y = 0$. Find the time when $v_0 \\sin\\theta - g t = 0$, then read $y$ at that instant."
        },
        {
          "md": "You should arrive at $H = \\dfrac{\\left(v_0 \\sin\\theta\\right)^2}{2g}$. Notice it depends only on the *vertical* launch speed — the horizontal motion never enters.",
          "fragment": true
        }
      ],
      "speaker_notes": "Tie the punchline back to slide 1: height is a purely vertical question, so $v_x$ cannot appear in the answer. Keep the fragment hidden until a group states that out loud."
    }
  ]
}