example_problem.json

{
  "example_problem": "Problem: A basketball is thrown with an initial velocity of 20 m/s at an angle of 45\u00b0 above the horizontal. The basketball passes through the center of a hoop before landing. If the basketball is moving horizontally at the instant it goes through the hoop, how high, H, above the ground is the center of the hoop?\nStep 1: Understand Problem\nWe need to find the height H of the basketball hoop. The basketball is thrown at an angle and passes through the hoop when it is moving horizontally. This means at the top of its trajectory, the vertical component of the velocity is zero.\nStep 2: Split into Components\nThe initial velocity has two components: horizontal and vertical. The horizontal component is \\( v_{0x} = v_0 \\cos \\theta \\) and the vertical component is \\( v_{0y} = v_0 \\sin \\theta \\).\nStep 3: Apply Relevant Equations\nAt the top of the trajectory, the vertical velocity is zero. Use the kinematic equation for vertical motion: \\( v_y^2 = v_{0y}^2 - 2gH \\), where \\( v_y = 0 \\).\nStep 4: Perform algebra and arithmetic\nSubstitute the components into the equation: \\( 0 = (20 \\sin 45\u00b0)^2 - 2 \\cdot 9.8 \\cdot H \\). Solve for H: \n\n \\( H = \\frac{(20 \\sin 45\u00b0)^2}{2 \\cdot 9.8} \\) \n\n Calculate: \n\n \\( H = \\frac{(20 \\cdot \\frac{\\sqrt{2}}{2})^2}{19.6} \\) \n\n \\( H = \\frac{(20 \\cdot 0.707)^2}{19.6} \\) \n\n \\( H = \\frac{(14.14)^2}{19.6} \\) \n\n \\( H = \\frac{200}{19.6} \\) \n\n \\( H \\approx 10.2 \\text{ meters} \\).\n"
}