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.

							
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
 
						

							
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.