Question

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

• A. 24.1 m , 19.6 m and 4.9 m above the top
• B. 44.1 m , 19.6 m and 4.9 m bellow the top
• C. 44.1 m , 12.6 m and 4.9 m bellow the top
• D. 41.4 m , 29.6 m and 4.9 m bellow the top

Answer 1

The correct option is B 44.1 m , 19.6 m and 4.9 m bellow the top

When the 6th ball is being dropped, the time passed since the balls have been dropped are-

3rd ball: 3 seconds

4th ball: 2 seconds

5th ball: 1 second

Hence the distance through which the balls fell in the time is:

3rd ball: $\frac{1}{2}g{t}^2=\frac{1}{2}\times 9.8\times 3^2$$=44.1m$

4th ball: $\frac{1}{2}g{t}^2=\frac{1}{2}\times 9.8\times 2^2$$=19.6m$

5th ball: $\frac{1}{2}g{t}^2=\frac{1}{2}\times 9.8\times 1^2$$=4.9m$