A cricketer of height $2.5 \mathrm{~m}$ throws a ball at an angle of $30^{\circ}$ with the horizontal such that it is received by another cricketer of same height standing at a distance of $50 \mathrm{~m}$ from the first one. The maximum height attained by the ball is ($\tan 30^{\circ}=0.577$)

The height of both cricketers is $h = 2.5 \text{ m}$. Since the ball is thrown and received at the same height, the vertical displacement is zero. The horizontal range $R$ is given by $R = \dfrac{u^2 \sin(2\theta)}{g}$.

Given $R = 50 \text{ m}$ and $\theta = 30^{\circ}$, we have $50 = \dfrac{u^2 \sin(60^{\circ})}{g} = \dfrac{u^2 \sqrt{3}}{2g}$.

Thus, $\dfrac{u^2}{g} = \dfrac{100}{\sqrt{3}}$.

The maximum height $H$ attained by the projectile above the point of projection is given by $H = \dfrac{u^2 \sin^2 \theta}{2g}$.

Substituting the value of $\dfrac{u^2}{g}$:

$H = \dfrac{1}{2} \times \left( \dfrac{u^2}{g} \right) \times \sin^2(30^{\circ}) = \dfrac{1}{2} \times \dfrac{100}{\sqrt{3}} \times \left( \dfrac{1}{2} \right)^2 = \dfrac{100}{8\sqrt{3}} = \dfrac{12.5}{1.732} \approx 7.216 \text{ m}$.

The total maximum height from the ground is $H_{total} = h + H = 2.5 + 7.216 = 9.716 \text{ m}$.

Rounding to the nearest value provided in the options, we get $9.7 \text{ m}$.

Answer: 9.7 m