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. 5 m
C. 7.5 m
D. 10 m

Explanation

To solve the problem of finding the maximum height attained by a ball thrown by a player, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Time of Flight**:
The time of flight (T) is given as 2 seconds.

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

3. **Rearrange the Formula**:
Rearranging the formula to find \( u \sin \theta \):
\[ u \sin \theta = \frac{gT}{2} \]
Substituting \( T = 2 \) seconds and \( g = 10 \, \text{m/s}^2 \):
\[ u \sin \theta = \frac{10 \times 2}{2} = 10 \, \text{m/s} \]

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

5. **Substitute \( u \sin \theta \)**:
We know \( u \sin \theta = 10 \, \text{m/s} \). To find \( u^2 \sin^2 \theta \), we square the previous result:
\[ u^2 \sin^2 \theta = (u \sin \theta)^2 = (10)^2 = 100 \, \text{m}^2/\text{s}^2 \]

6. **Calculate Maximum Height**:
Now substitute \( u^2 \sin^2 \theta \) into the maximum height formula:
\[ H = \frac{100}{2 \times 10} = \frac{100}{20} = 5 \, \text{m} \]

7. **Final Answer**:
The maximum height attained by the ball above the point of projection is **5 meters**.