A ball is dropped downwards. After 1 second another ball is dropped downwards from the same point. What is the distance between them after 3 seconds

Options

A. 25m
B. 20m
C. 50m
D. 9.8m

Answer

To solve the problem, we need to analyze the motion of both balls separately and then find the distance between them after 3 seconds.

### Step-by-Step Solution:

1. **Identify the motion of the first ball:**
- The first ball is dropped from rest. It falls for a total of 3 seconds.
- The displacement \( S_1 \) of the first ball can be calculated using the formula:
\[ S_1 = ut + \frac{1}{2}gt^2 \]
where \( u = 0 \) (initial velocity), \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)), and \( t = 3 \, \text{s} \).
- Thus:
\[ S_1 = 0 + \frac{1}{2} \cdot g \cdot (3)^2 = \frac{1}{2} \cdot 10 \cdot 9 = 45 \, \text{m} \]

2. **Identify the motion of the second ball:**
- The second ball is dropped 1 second after the first ball, so it falls for a total of 2 seconds.
- The displacement \( S_2 \) of the second ball can be calculated similarly:
\[ S_2 = ut + \frac{1}{2}gt^2 \]
where \( t = 2 \, \text{s} \).
- Thus:
\[ S_2 = 0 + \frac{1}{2} \cdot g \cdot (2)^2 = \frac{1}{2} \cdot 10 \cdot 4 = 20 \, \text{m} \]

3. **Calculate the distance between the two balls:**
- The distance between the two balls after 3 seconds is the difference in their displacements:
\[ \text{Distance} = S_1 - S_2 = 45 \, \text{m} - 20 \, \text{m} = 25 \, \text{m} \]

### Final Answer:
The distance between the two balls after 3 seconds is **25 meters**.