Question

{"given":"Initial speed u = 10 m/s, angle of projection θ = 30°, height of building = 10 m, acceleration due to gravity g = 10 m/s², sin 30° = 1/2, cos 30° = √3/2. The ball is thrown from a height of 10 m and we need to find the horizontal distance when it returns to the same height of 10 m from ground.","key_observation":"When the ball returns to the same height from which it was thrown (10 m from ground), it completes its full trajectory. The horizontal distance covered during this time is the maximum horizontal range for projectile motion. This occurs because the ball starts and ends at the same vertical level, making the time of flight equal to the complete parabolic trajectory time.","option_analysis":[{"label":"(A)","text":"$5.20$ m","verdict":"incorrect","explanation":"This value is less than the calculated range. It might represent an intermediate calculation or a partial horizontal distance, but not the complete range when the ball returns to the initial height."},{"label":"(B)","text":"$4.33$ m","verdict":"incorrect","explanation":"This is approximately half of the correct answer (8.66/2 = 4.33). This might be the horizontal distance to reach maximum height, not the complete range to return to the same level."},{"label":"(C)","text":"$2.60$ m","verdict":"incorrect","explanation":"This value is significantly smaller than the correct range. It represents approximately 30% of the actual range and doesn't correspond to any standard projectile motion calculation for this problem."},{"label":"(D)","text":"$8.66$ m","verdict":"correct","explanation":"Step 1: For maximum horizontal range formula:\n$$R = \\frac{u^2 \\sin 2\\theta}{g}$$\nStep 2: Substitute the values:\n$$R = \\frac{10^2 \\sin(2 \\times 30°)}{10} = \\frac{100 \\sin 60°}{10}$$\nStep 3: Calculate using sin 60° = √3/2:\n$$R = \\frac{100 \\times \\frac{\\sqrt{3}}{2}}{10} = 5\\sqrt{3} = 5 \\times 1.732 = 8.66 \\text{ m}$$"}],"answer":"(D)","formula_steps":[]}