A ball thrown by one player reaches the other in `2 s`. The maximum height attained by the ball above the point of projection will be about.

Options

A. `2.5 m`
B. `5m`
C. `7.5 m`
D. `10m`

Explanation

To find the maximum height attained by the ball above the point of projection, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**:
The ball is thrown and reaches another player in 2 seconds. We need to determine the maximum height it reaches during its flight.

2. **Identify the Time of Flight**:
The total time of flight (T) for a projectile is given as 2 seconds.

3. **Use the Time of Flight Formula**:
The time of flight for a projectile is given by the formula:
\[ T = \frac{2u \sin \theta}{g} \]
where \( u \) is the initial velocity, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)).

4. **Set Up the Equation**:
Since the time of flight is 2 seconds, we can set up the equation:
\[ 2 = \frac{2u \sin \theta}{g} \]

5. **Rearranging the Equation**:
Rearranging gives us:
\[ u \sin \theta = g \]

6. **Maximum Height Formula**:
The maximum height (H) attained by a projectile is given by:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]

7. **Substituting for \( u \sin \theta \)**:
From the previous step, we know that \( u \sin \theta = g \). Therefore, we can substitute \( g \) into the height formula:
\[ H = \frac{(u \sin \theta)^2}{2g} = \frac{g^2}{2g} \]

8. **Simplifying the Expression**:
Simplifying gives us:
\[ H = \frac{g}{2} \]

9. **Substituting the Value of \( g \)**:
Using \( g = 10 \, \text{m/s}^2 \):
\[ H = \frac{10}{2} = 5 \, \text{m} \]

10. **Conclusion**:
The maximum height attained by the ball above the point of projection is **5 meters**.