A boy learning cricket finds that the greatest distance he can bowl the ball is $36m$. If the maximum height of the trajectory (in a given scenario) is $n\,metres$, find value of $n$. [Neglect air resistance].

Hint: The ball follows projectile motion, the distance covered in the horizontal direction is called range while the distance covered in the vertical direction is the height. In the vertical direction force of gravity acts on the ball while no force acts on the horizontal direction. As the acceleration is constant, we can use equations of motion in the vertical direction.

Complete step by step answer:
Let the velocity of the ball be $v$. Since, the ball follows a projectile path; its velocity will have a horizontal component as well as a vertical component. Its horizontal component is given by $v\cos \theta $ and its vertical component is given by $v\sin \theta $.
The range is the distance covered along the ground. The range will have the maximum value when$\theta ={{45}^{o}}$.
The ball is under the action of acceleration due to gravity in the vertical direction and at the highest point the velocity becomes zero. Applying the following equation of motion,
${{v}^{2}}={{u}^{2}}+2as$
Here, $v$ is the final velocity
$u$ is the initial velocity
$a$ is the acceleration
$s$ is the distance travelled
In the above equation, we substitute given values to get,
$\begin{align}
  & 0={{(v\sin 45)}^{2}}-2gs \\
 & \Rightarrow 0={{\left( \dfrac{v}{\sqrt{2}} \right)}^{2}}-2\times 10\times s \\
 & \Rightarrow \dfrac{{{v}^{2}}}{2}=20s \\
\end{align}$
$\therefore s=\dfrac{{{v}^{2}}}{40}m$ - (1)
Using the following equation of motion,
$v=u+at$
Here, $t$ is the time taken
$\begin{align}
  & 0=v\sin 45-gt \\
 & \Rightarrow \dfrac{v}{g\sqrt{2}}=t \\
\end{align}$
Therefore, the time taken by the ball to cover the range will be $2t=2\dfrac{v}{g\sqrt{2}}$
The horizontal component remains constant throughout therefore,
$v\cos 45=\dfrac{R}{2t}$
Here, $R$ is the range
Given, $R=36m$ substituting given values in the above equation we get,
$\begin{align}
  & \dfrac{v}{\sqrt{2}}=\dfrac{36}{2\times \dfrac{v}{10\times \sqrt{2}}} \\
 & \Rightarrow {{v}^{2}}=360m \\
 & \Rightarrow v=6\sqrt{10}m \\
\end{align}$
Substituting the value of velocity in eq (1) we get,
$\begin{align}
  & s=\dfrac{{{v}^{2}}}{40}m \\
 & \Rightarrow s=\dfrac{360}{40} \\
 & \therefore s=9m \\
\end{align}$

Therefore, the maximum height of the trajectory is $9m$. Hence, the value of $n$ is $9$.

Note: At the highest point, the potential energy is maximum and the kinetic energy is zero so the velocity is zero. While at the lowest point, the kinetic energy is maximum and the potential energy is minimum. No acceleration acts on the horizontal component of a body in projectile motion.