Slide down and go the distance

Let the velocity the ball leaving the ramp be $v \text{ ms}^{-1}$ . Since the ball leaves horizontally, the horizontal and vertical components of $v$ are $v_x = v$ and $v_y = 0$ respectively.

The time $t$ taken for the ball to drop a distance of $h$ is given by:

$\begin{aligned} h & = v_yt + \frac 12 gt^2 \\ 2 & = 0 + \frac 12 (10)t^2 \\ \implies t & = \sqrt{\frac 25} \text{ s} \end{aligned}$

In time $t$ , the horizontal distance travelled by the ball $d$ is given by:

$\begin{aligned} d & = v_x t \\ \implies v_x & = \frac dt \\ v & = 4 \times \sqrt{\frac 52} = \sqrt{40} \text{ ms}^{-1} \end{aligned}$

When the ball drop from $H$ to $h$ , its potential energy is converted to kinetic energy as follows:

$\begin{aligned} mg(H-h) & = \frac 12 mv^2 \\ \implies H & = \frac {v^2}{2g} + h = \frac {40}{2(10)} + 2 = \boxed{4} \text{ m} \end{aligned}$

Relevant wiki: Understanding the work-kinetic energy theorem

Applying Work-Energy Theorem between points A and B $\begin{aligned} mgH &= mg h_1 +\frac{1}{2}m v^2 \\ v^2 +2g H_1 &=2gH & (1) \\ \end{aligned}$ Since particle strikes the ground at a horizontal distance of 4 m , $\begin{aligned} v \times \sqrt{\frac{2 H_1}{g}} &=d \\ v^2 & = \frac{gd^2}{2 H_1} & (2) \\ \end{aligned}$ Substituting value of $v^2$ from equation (2) in equation (1) , We get $\begin{aligned} 2gH &= \frac{gd^2}{2 H_1} + 2gH_1 \\ \end{aligned}$ Putting $H_1= \SI{2}{\meter}$ , $d= \SI{4}{\meter}$ , and $g= \SI[per-mode=symbol]{10}{\meter\per\second\squared},$ we get, $\large{H} =\boxed{\SI[per-mode=symbol]{4}{\meter}}.$