Question

a body is droped from rest from height h.it covers9/25h distance iin the last second of the fall. find h

Answer 1

Let total time taken be T. According to Newton's second law: S = UT + 1/2AT² where, S represents total distance, U represents initial velocity, A represents acceleration due to gravity which is 9.8m/s² and T represents time taken. Since, the object is dropped from the top of the building, initial velocity is zero, so the value of UT is 0. Hence, the equation can be rewritten as S = 1/2AT². Let the total time taken to drop to the ground be T and the time taken prior to the last second be T-1. Hence, S = 1/2AT² and 16/25S = 1/2A(T-1)². Substituting the value of A as 9.8m/s², we get S = 1/2*9.8*T² which gives, S = 4.9T². substituting the value of S = 4.9T² in the second equation, we get 16/25*4.9T² = 1/2*9.8*(T-1)². Hence, 16/25*4.9T² = 4.9*(T-1)². Now, cancel 4.9 in both the sides of the equation. Hence, we get 16/25T² = (T-1)². Now we get, 16/25T² = T² - 2T +1. By solving, we get, 2T-1 = T² - 16/25T². Hence, by solving, we get 9/25T² - 2T + 1 = 0. Multiplying both sides by 25, we get, 9T² - 50T + 25 = 0. Now, we have to solve using quadratic formula which is (-b+/- {sq. Root b² - 4ac})/ 2a where a = 9, b = -50 and c = 25. Solving on the above equation, we get 2 solutions for T, where T = 5 (or) T = 5/9. Since, based on the question, time taken is more than one second. Hence, the second solution where T = 5/9 is not feasible since it is lesser than 1. Hence, the correct solution is T = 5. Hence, the height of the building is 1/2*9.8*5² which comes to 122.5 metres. Hence, the height of the building is 122.5 metres :)