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Grade: 12

                        

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

Answers : (2)

Pratham Ashish
17 Points
							

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
Umar Iliyas Bhokare
15 Points
							
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.
2 years ago
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