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A stone is dropped from rest from the top of tower 19.6m high . Find the distance covered during the last second of its fall |
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Answer» Answer: s=14.7 m Explanation: We can use the formula, [s (nth second) = u + 1/2 a (2 N - 1)], (Where u = initial velocity a = acceleration and s= displacement) which can easily be derived by subtracting the displacement in (n - 1) SECONDS from the total displacement in 'n' seconds. (Kinematical equation for displacement in 'n' seconds is s= un + 1/2 a n²) Total TIME taken for stone to fall, (Here acceleration = G = 9.8 m/s²) 19.6 = (0)t + 1/2(g)n² => n² = (19.6 x 2)/ 9.8 => n² = 4 => n = 2 seconds (total time taken for stone to fall) Therefore, displacement in last, that is the 2nd second, S = (0) + 1/2(g) (2(2) - 1) (Here u = 0, n = 2 and acceleration = g) => s = 1/2 x 9.8 x 3 => s = 14.7 m <-- Ans. |
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