SOLUTION: a body falls freely from the top of the toeer and during last second of the fall ,it falls through 25m.Find the height of tower.

Question 1033304: a body falls freely from the top of the toeer and during last second of the fall ,it falls through 25m.Find the height of tower.
Found 2 solutions by Alan3354, ikleyn:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
a body falls freely from the top of the toeer and during last second of the fall ,it falls through 25m.Find the height of tower.
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You have to spec the acceleration of gravity.
Answer by ikleyn(52786) (Show Source): You can put this solution on YOUR website!
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A body falls freely from the top of the tower and during last second of the fall, it falls through 25 m. Find the height of tower.
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Solution 1

This fact is known very well from Physics, or from Calculus or from . . . Free falling body falls the distance in t seconds, where "g" is the gravity acceleration. Therefore, your equation to find the time is - = 25. In this problem, take g = 10 (actually, g = 9.81 ). You will get - = 25, or = 25, or = 25, or 10t = 30. Hence, t = 3 seconds. For 3 seconds, the body falls = = 5*9 = 45 m. Answer. The height of the building is 45 m.

Solution 2

Calculate the distances the free falling body falls during the 1-st, 2-nd, 3-rd . . . seconds. Use g = 10 . 1-st sec.: = = 5 m. 2-nd sec.: = = 20-5 = 15 m. 3-rd sec.: = = 45-20 = 25 m. See these numbers: 5, 15, 25 . . . They form ARITHMETIC PROGRESSION !!! This remarkable fact is general: The distances that the free falling body falls during the first second, the next one, the third and so on, form the arithmetic progression. Miracle ?! - No, the algebra only. - See the lessons Free fall and arithmetic progressions Uniformly accelerated motions and arithmetic progressions in this site. And not to forget, the calculations above that lead to the number "25 m in third second" give another solution to the original problem.

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