A stone is dropped from a tall tower. The distance it falls in the last second of its motion is exactly equal to the total distance it falls in the first 3 seconds. What is the total height of the tower? (Take g = 10 m/s ^2 )

Step-by-step solution

Let the total time of fall be n seconds. The total distance fallen in the first 3 seconds is: S₃ = 1 2 gt^2 = 1 2 10 (3)^2 = 5 9 = 45 m The distance fallen in the last ( n^ th ) second is given by: S_n = u + g 2 (2n - 1) Since the stone is dropped, u = 0 . S_n = 10 2 (2n - 1) = 5(2n - 1) According to the problem, the distance in the last second equals the distance in the first 3 seconds: 5(2n - 1) = 45 2n - 1 = 9 2n = 10 n = 5 s The total time of flight is 5 seconds. The total height of the tower is the total dista