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.

Explanation

To solve the problem step by step, we can follow these steps:

### Step 1: Understand the problem
We need to find the maximum height attained by a ball thrown by one player that reaches another player in 2 seconds.

### Step 2: Use the time of flight formula
The time of flight \( T \) for a projectile is given by the formula:
\[ T = \frac{2u \sin \theta}{g} \]
where:
- \( T \) is the time of flight (2 seconds in this case),
- \( u \) is the initial velocity,
- \( \theta \) is the angle of projection,
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).

### Step 3: Rearranging the formula
From the time of flight formula, we can rearrange it to find \( u \sin \theta \):
\[ u \sin \theta = \frac{gT}{2} \]
Substituting \( T = 2 \) seconds:
\[ u \sin \theta = \frac{g \cdot 2}{2} = g \]

### Step 4: Substitute the value of \( g \)
Using \( g \approx 9.81 \, \text{m/s}^2 \):
\[ u \sin \theta = 9.81 \, \text{m/s} \]

### Step 5: Use the maximum height formula
The maximum height \( H \) attained by the projectile is given by:
\[ H = \frac{(u \sin \theta)^2}{2g} \]
Substituting \( u \sin \theta = 9.81 \, \text{m/s} \):
\[ H = \frac{(9.81)^2}{2 \cdot 9.81} \]

### Step 6: Simplify the equation
This simplifies to:
\[ H = \frac{96.2361}{19.62} \approx 4.9 \, \text{m} \]

### Step 7: Conclusion
Thus, the maximum height attained by the ball above the point of projection is approximately \( 4.9 \, \text{m} \).