{
  "schema_version": 1,
  "prompt": {
    "type": "doc",
    "blocks": [
      {
        "type": "paragraph",
        "content": [
          {
            "type": "text",
            "text": "A ball is launched from ground level over flat terrain with speed "
          },
          { "type": "math", "latex": "v_0 = 20\\,\\text{m/s}" },
          { "type": "text", "text": " at an angle of " },
          { "type": "math", "latex": "45^\\circ" },
          {
            "type": "text",
            "text": " above the horizontal. Ignoring air resistance and taking "
          },
          { "type": "math", "latex": "g = 9.8\\,\\text{m/s}^2" },
          {
            "type": "text",
            "text": ", how far from the launch point does it land, in metres?"
          }
        ]
      },
      {
        "type": "callout",
        "variant": "info",
        "blocks": [
          {
            "type": "paragraph",
            "content": [
              {
                "type": "text",
                "text": "Give your answer to one decimal place. A tolerance of 0.5 m is accepted."
              }
            ]
          }
        ]
      }
    ]
  },
  "answer": {
    "type": "numeric",
    "value": 40.8,
    "tolerance": { "type": "absolute", "value": 0.5 },
    "unit": "m"
  },
  "hints": [
    {
      "type": "doc",
      "blocks": [
        {
          "type": "paragraph",
          "content": [
            {
              "type": "text",
              "text": "The trajectory is a downward-opening parabola. The range of an ideal projectile launched from ground level is"
            }
          ]
        },
        {
          "type": "math_block",
          "latex": "R = \\frac{v_0^2 \\sin(2\\theta)}{g}."
        }
      ]
    }
  ],
  "solution": {
    "type": "doc",
    "blocks": [
      {
        "type": "paragraph",
        "content": [
          { "type": "text", "text": "With " },
          { "type": "math", "latex": "\\theta = 45^\\circ" },
          { "type": "text", "text": ", we have " },
          { "type": "math", "latex": "\\sin(2\\theta) = \\sin 90^\\circ = 1" },
          { "type": "text", "text": ", so" }
        ]
      },
      {
        "type": "math_block",
        "latex": "R = \\frac{v_0^2}{g} = \\frac{20^2}{9.8} = \\frac{400}{9.8} \\approx 40.8\\,\\text{m}."
      },
      {
        "type": "paragraph",
        "content": [
          {
            "type": "text",
            "text": "This is also why 45 degrees maximises range on level ground: it maximises the quadratic factor "
          },
          { "type": "math", "latex": "\\sin(2\\theta)" },
          { "type": "text", "text": "." }
        ]
      }
    ]
  }
}