 # 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.

13 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

4 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)meters
Hence,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.