Two balls are released from the same height at an interval of 2s. When will the separation between the balls be 20m after the first ball is released? Take `g=10m//s^(2)`.

Answer

To solve the problem of finding when the separation between two balls released from the same height at an interval of 2 seconds will be 20 meters, we can follow these steps:

### Step-by-Step Solution

1. **Understanding the Problem**:
- Two balls are released from the same height.
- The first ball (Ball A) is released at time \( t = 0 \).
- The second ball (Ball B) is released at time \( t = 2 \) seconds.
- We need to find the time \( t \) after the first ball is released when the separation between the two balls is 20 meters.

2. **Using the Equation of Motion**:
- The distance fallen by an object under free fall can be calculated using the second equation of motion:
\[ s = ut + \frac{1}{2} g t^2 \]
- Here, \( u \) (initial velocity) is 0 (since the balls are released), \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity), and \( t \) is the time in seconds.

3. **Calculating the Distance Fallen by Ball A**:
- For Ball A, which has fallen for time \( t \):
\[ s_A = 0 + \frac{1}{2} \cdot 10 \cdot t^2 = 5t^2 \]

4. **Calculating the Distance Fallen by Ball B**:
- For Ball B, which starts falling 2 seconds after Ball A, the time it has been falling when Ball A has been falling for \( t \) seconds is \( t - 2 \) seconds (only if \( t \geq 2 \)):
\[ s_B = 0 + \frac{1}{2} \cdot 10 \cdot (t - 2)^2 = 5(t - 2)^2 \]

5. **Finding the Separation Between the Balls**:
- The separation \( d \) between the two balls is given by:
\[ d = s_A - s_B = 5t^2 - 5(t - 2)^2 \]
- Expanding \( (t - 2)^2 \):
\[ (t - 2)^2 = t^2 - 4t + 4 \]
- Thus, we have:
\[ d = 5t^2 - 5(t^2 - 4t + 4) = 5t^2 - 5t^2 + 20t - 20 = 20t - 20 \]

6. **Setting Up the Equation**:
- We want the separation \( d \) to be 20 meters:
\[ 20t - 20 = 20 \]

7. **Solving for \( t \)**:
- Rearranging the equation:
\[ 20t = 40 \implies t = 2 \]

8. **Conclusion**:
- The separation between the two balls will be 20 meters after **2 seconds** from the release of the first ball.