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

To find the maximum height attained by the ball, we can analyze its vertical motion.

The total time of flight of the ball from one player to the other is given as:
$T=2\text{s}$

Assuming the point of projection and the point of catch are at the same horizontal level, the time taken by the ball to reach its maximum height (time of ascent, $t_a$) is exactly half of the total time of flight:
$t_a=\frac{T}{2}=\frac{2}{2}=1\text{s}$

At the maximum height, the vertical component of the velocity ($v_y$) becomes zero.
Using the first equation of motion for the upward journey:
$v_y=u_y-g{t}_a$
where $u_y$ is the initial vertical velocity and $g$ is the acceleration due to gravity (approximately $9.8{\text{m/s}}^2$ or $10{\text{m/s}}^2$).

Setting $v_y=0$ and using $g\approx 10{\text{m/s}}^2$:
$0=u_y-\left(10\right)\left(1\right)$
$u_y=10\text{m/s}$

Now, we can find the maximum height ($H$) using the second equation of motion for the upward journey:
$H=u_y{t}_a-\frac{1}{2}g{t}_a^2$
Substitute the values:
$H=\left(10\right)\left(1\right)-\frac{1}{2}\left(10\right){\left(1\right)}^2$
$H=10-5=5\text{m}$

Therefore, the maximum height attained by the ball above the point of projection is about 5 m.