Umar Iliyas Bhokare
Last Activity: 6 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.