A stone is dropped from a tower. The distance covered by it in the last second of its fall is equal to the distance covered in the first three seconds. The height of the tower is approximately: (g = 10 m/s²)
Explanation:
Distance in first 3 seconds: s₃ = ½gt² = ½×10×9 = 45 m. Let total time be n seconds. Distance in nth second: sₙ = u + g(2n-1)/2 = 0 + 5(2n-1) = 10n - 5. Given sₙ = 45 m ⇒ 10n - 5 = 45 ⇒ n = 5 s. Total height h = ½gn² = ½×10×25 = 125 m. This problem combines equation of motion with logical reasoning about time intervals. Exam tip: For free fall from rest, distance in nth second = 5(2n-1) meters when g=10 m/s². Such numerical problems test application skills in competitive exams.
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