Where is the root coming from u should be going step by step I can't understand that

The initial potential energy of $M_2$ equals the final potential energy of $M_1$ plus the final kinetic energy of both masses.

$$M_2 g h_2 = M_1 g h_2 + \frac{1}{2} M_1 v_f^2 + \frac{1}{2} M_2 v_f^2$$

Subtract $M_1 g h_2$ from both sides:

$$M_2 g h_2 - M_1 g h_2 = \frac{1}{2} M_1 v_f^2 + \frac{1}{2} M_2 v_f^2$$

On the left side, $g h_2$ is common:

$$g h_2 (M_2 - M_1) = \frac{1}{2} M_1 v_f^2 + \frac{1}{2} M_2 v_f^2$$

On the right side, $\frac{1}{2} v_f^2$ is common:

$$g h_2 (M_2 - M_1) = \frac{1}{2} v_f^2 (M_1 + M_2)$$

To get $v_f^2$ by itself, first multiply both sides by 2:

$$2 g h_2 (M_2 - M_1) = v_f^2 (M_1 + M_2)$$

Now, divide both sides by $(M_1 + M_2)$:

$$\frac{2 g h_2 (M_2 - M_1)}{M_1 + M_2} = v_f^2$$

Since $v_f^2$ is equal to the expression, $v_f$ is the square root of that expression:

$$v_f = \sqrt{\frac{2 g h_2 (M_2 - M_1)}{M_1 + M_2}}$$

This is where the square root comes from. It's the final step in isolating $v_f$ from $v_f^2$.

Now, let's re-substitute the values:
$M_1 = 1.5 \text{ kg}$
$M_2 = 4.5 \text{ kg}$
$h_2 = 2 \text{ m}$
$g = 9.81 \text{ m/s}^2$

$$v_f = \sqrt{\frac{2 \times 9.81 \text{ m/s}^2 \times 2 \text{ m} \times (4.5 \text{ kg} - 1.5 \text{ kg})}{1.5 \text{ kg} + 4.5 \text{ kg}}}$$ $$v_f = \sqrt{\frac{2 \times 9.81 \times 2 \times 3.0}{6.0}} \text{ m/s}$$ $$v_f = \sqrt{\frac{117.72}{6.0}} \text{ m/s}$$ $$v_f = \sqrt{19.62} \text{ m/s}$$ $$v_f \approx 4.429 \text{ m/s}$$

The velocity with which $M_2$ hits the ground is approximately $\boxed{\text{4.43 m/s}}$.