Question

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.

Answer 1

A person is releasing balls from a tall building at regular intervals of one second.
It means for each ball, the initial velocity u is 0.
Acceleration due to gravity, a = g = 9.8 m/s²
When the 6^th ball is dropped, the 5^th ball moves for 1 second, the 4^th ball moves for 2 seconds and the 3^rd ball moves for 3 seconds.
Position of the 3^rd ball after t = 3 s:
Using the equation of motion, we get:

\[s_3 = ut + \frac{1}{2}a t^2\]

\[\Rightarrow s_3 = 0 + \frac{1}{2} \times 9 . 8 \times 3^2 = 44 . 1 \text{ m } \]

(from the top of the building)

Position of the 4^th ball after t = 2 s:

\[s_4 = ut + \frac{1}{2}a t^2\]

\[\Rightarrow s_4 = 0 + \frac{1}{2} \times 9 . 8 \times 2^2 = 19 . 6 \text{ m } \]

(from the top of the building)

Position of the 5^th ball after t = 1 s:

\[s_5 = ut + \frac{1}{2}a t^2\]

\[\Rightarrow s_5 = 0 + \frac{1}{2} \times 9 . 8 \times 1^2 = 4 . 9 \text{ m } \]

(from the top of the building)