Question 1033304
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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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<U>Solution 1</U>


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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.
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<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.
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