Lesson The distance covered by a free falling body during last second of the fall

The distance covered by a free falling body during last second of the fall

Problem 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.

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.