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A person sitting on the top of a tall building is dropping balls at regular intervals of one second.Find the positions of the 3rd,4th and 5th ball when the 6th ball is being dropped.

```
11 years ago

```							When the sixth ball was dropped, 5 seconds have elapsed from the time the first ball was dropped, 4 seconds after the second ball was dropped, 3 seconds after the the third ball was  dropped, 2 seconds after the fourth ball was dropped and 1 second  after the fifth ball was dropped. Thus being said, when the sixth ball was dropped & since the 3rd ball has been falling for 3 seconds, then D3 = (1/2)(9.8)(3^2) = 44.1 meters Since the fourth ball has been falling for 2 seconds, D4 = (1/2)(9.8)(2^2) = 19.6 meters and since the fifth ball has been falling for 1 second, D5 = (1/2)(9.8)(1)^2 = 4.9 m

```
11 years ago
```							When the sixth ball was dropped,the initial velocity will be zero.Considering the time taken by the 3rd,4th and the 5th ball as 1/3,1/2 and 1,we can easily find the height of 3rd ball (H1) from the top of building ,height of 4th ball (H2) and height of 5th ball (H3). Here, u=0, a=(9.8) m/s By,    S=ut+1/2at^2,                   {ut is eliminated because u =0}  1. -H1 = –½ x (9.8) x 9       (H1)= ½ x 88.2             =(44.1)meters   2. -H2 = – ½ x (9.8) x 2      (H2)= ½ x 19.6            =(9.8)meters  3. -H3 = – ½ x (9.8) x 1     (H3)= ½ x 9.8           =(4.9)metersHence,when the 6th ball is being dropped,the position of the 3rd,4th,5th balls from the top of the building are 44.1m,19.6m and 4.9m respectively.
```
2 years ago
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