You really need to be careful with this...

same problem as me.. I approach the answer must 1.2755...

You're assuming the block will undergo projectile motion from the bottom of the wedge itself! But that's not possible (block can't go through the wedge). So, first block will reach top of the wedge and then projectile motion (free fall).

You made a mistake. What Ritesh Yadav did is correct.

Kinetic Energy is a scalar quantity. The $v$ in $\dfrac{1}{2}mv^2$ doesn't care which way your velocity is pointing. That's the whole allure of the energy approach—we can ignore direction and just do our calculations solely using scalars and still get the right answer.

Let's think about the block. It's on a wedge which doesn't move throughout the scenario. You correctly said that gravity is acting on the object. What you missed is the fact that the wedge is also exerting a normal force perpendicular to the its surface. When we resolve this along the $x-$ and $y-$ directions, you'll find that the acceleration in the $y$ -direction isn't exactly $g$ . Moreover, you'll find that the block's velocity's $x$ -component changes too.

This makes sense. Think about it. The block's velocity's $x$ -component has to proportionally change with the $y-$ component. Otherwise, the angle that the velocity makes with the $x-$ axis won't be 30 degrees and the block will either start flying before it reaches the end of the wedge or go through the wedge—both of which are nonsensical scenarios.

Let's do the calculation your way, taking the normal force exerted by the wedge on the block into account. (Just note that in the wild, one would be heavily discouraged from trying to do it this way since it's so calculation-intensive and unnecessary.)

The net force in the $y$ -direction, $F_y$ , is $-mg + N\cos\theta$ where $g$ is the acceleration due to gravity, $N$ is the normal force, and $\theta$ is the angle that the wedge makes with the horizontal, which is 30 degrees. (I'm taking the downward direction as negative.)

Similarly, the net force in the $x-$ direction is $F_x = -N\sin\theta$ .

We know that $N = mg\cos\theta$ , which means that $a_y = -g(1 - \cos^2\theta ) = -g\sin^2\theta = -\dfrac{1}{4}g$ and that $a_x = -g\sin\theta\cos\theta = -\dfrac{\sqrt{3}}{4}g$

(For the problem, we take $g = 10 \ \mathrm{m\cdot s^{-2}}$ .)

If we use these values in the equation $v^2 = u^2 + 2as$ (note that $a$ and $s$ need to be in the same direction), then we get $v_{y/\text{top}} = \sqrt{5^2 - 2\left(\dfrac{1}{4}g\right)(1.8)} = 4\ \mathrm{m\cdot s^{-2}}$ and $v_{x/\text{top}} = \sqrt{\big(5\sqrt{3}\big)^2 - 2\left(\dfrac{\sqrt{3}}{4}g\right)(1.8\cot\theta)} = 4\sqrt{3} \ \mathrm{m\cdot s^{-2}}$ where $v_{y/\text{top}}$ is the $y-$ component of the block at the top of the wedge and $v_{x/\text{top}}$ is the $x-$ component of the block at the top of the wedge.

Strictly speaking, we don't need $v_{x/\text{top}}$ to do the problem. However, note that the angle made by the block's velocity with the horizontal is still $\tan^{-1}\left(\dfrac{4}{4\sqrt{3}}\right) = 30^{\circ}$ , exactly as we would expect.

Now that we have $v_{y/\text{top}}$ , we can use the equation $v^2 = u^2 + 2as$ once again to find the extra height traveled: $0 = 16 - 2(10)h \implies h = 0.8 \ \mathrm{m}$

Hence, the height attained by the block is $H + h = 2.6 \ \mathrm{m}$ .

TL;DR: You forgot to take the normal force into account.