Lesson The distance covered by a free falling body during last second of the fall
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<H2>The distance covered by a free falling body during last second of the fall</H2> <H3>Problem 1</H3>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. <U>Solution 1</U> <pre> This fact is known very well from Physics, or from Calculus or from . . . Free falling body falls the distance {{{(gt^2)/2}}} in t seconds, where "g" is the gravity acceleration. Therefore, your equation to find the time is {{{(g*t^2)/2}}} - {{{(g*(t-1)^2)/2}}} = 25. In this problem, take g = 10 {{{m/s^2}}} (actually, g = 9.81 {{{m/s^2}}}). You will get {{{(10*t^2)/2}}} - {{{(10*(t-1)^2)/2}}} = 25, or {{{5t^2 - 5(t-1)^2}}} = 25, or {{{5t^2 - 5t^2 + 10t - 5}}} = 25, or 10t = 30. Hence, t = 3 seconds. For 3 seconds, the body falls {{{(g*t^2)/2}}} = {{{(10*3^2)/2}}} = 5*9 = 45 m. <U>Answer</U>. The height of the building is 45 m. </pre> <U>Solution 2</U> <pre> Calculate the distances the free falling body falls during the 1-st, 2-nd, 3-rd . . . seconds. Use g = 10 {{{m/s^2}}}. 1-st sec.: {{{(gt^2)/2)}}} = {{{(10*1^2)/2}}} = 5 m. 2-nd sec.: {{{(g*2^2)/2 - (g*1^2)/2}}} = {{{(10*2^2)/2) - ((10*1^2)/2)}}} = 20-5 = 15 m. 3-rd sec.: {{{(g*2^3)/2 - (g*2^2)/2}}} = {{{(10*3^2)/2) - ((10*2^2)/2)}}} = 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 <A HREF=https://www.algebra.com/algebra/homework/Sequences-and-series/Free-fall-and-arithmetic-progressions.lesson>Free fall and arithmetic progressions</A> <A HREF=https://www.algebra.com/algebra/homework/Sequences-and-series/Uniformly-accelerated-motions-and-arithmetic-progressions.lesson>Uniformly accelerated motions and arithmetic progressions</A> 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. </pre> My other lessons on <B>Travel and Distance</B> problems in this site are <TABLE> <TR> <TD> - <A HREF=http://www.algebra.com/algebra/homework/word/travel/Travel-and-Distance-problems.lesson>Travel and Distance problems</A> - <A HREF=https://www.algebra.com/algebra/homework/word/travel/Travel-and-Distance-problems-for-two-bodies-moving-toward-each-other.lesson>Travel and Distance problems for two bodies moving in opposite directions</A> - <A HREF=https://www.algebra.com/algebra/homework/word/travel/Typical-catching-up-Travel-and-Distance-problems.lesson>Travel and Distance problems for two bodies moving in the same direction (catching up)</A> - <A HREF=http://www.algebra.com/algebra/homework/NumericFractions/Using-fractions-to-solve-Travel-problems.lesson>Using fractions to solve Travel problems</A> - <A HREF=http://www.algebra.com/algebra/homework/word/travel/Wind-and-Current-problems.lesson>Wind and Current problems</A> - 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