A stone is dropped from a very high cliff. If the distance covered by the stone in the n^ th second of its descent is exactly equal to the total distance covered by it in the first 3 seconds of its fall, what is the value of n ?

Step-by-step solution

Distance covered in the n^ th second is given by S_n = u + 1 2 g(2n - 1) . Since the stone is dropped, u = 0 . S_n = 1 2 g(2n - 1) Total distance covered in the first 3 seconds is given by S = ut + 1 2 gt^2 . S = 0 + 1 2 g(3)^2 = 9 2 g Given that S_n = S , we have: 1 2 g(2n - 1) = 9 2 g 2n - 1 = 9 2n = 10 n = 5 Answer: 5