* **Step 1: Identify the relevant information.**
We are given the following information:
* The initial velocity of the puck is 28.0 m/s horizontally.
* The height of the platform above the ground is 2.00 m.
* We can neglect air resistance.
* **Step 2: Set up the equations.**
We can use the following equations to solve for the direction of the velocity of the puck just before it hits the ground:
* $v_y = v_0y + at$
* $y = y_0 + v_0yt + \frac{1}{2}at^2$
Where:
* $v_y$ is the final velocity in the vertical direction
* $v_0y$ is the initial velocity in the vertical direction
* $a$ is the acceleration due to gravity (9.8 m/s^2)
* $t$ is the time
* $y$ is the vertical position
* $y_0$ is the initial vertical position
* **Step 3: Solve for the unknowns.**
We can solve for the final velocity in the vertical direction using the following equation:
```
v_y = v_0y + at
```
Plugging in the known values, we get:
```
v_y = 0 + (-9.8) t
```
```
v_y = -9.8t
```
We can solve for the time using the following equation:
```
y = y_0 + v_0yt + \frac{1}{2}at^2
```
Plugging in the known values, we get:
```
2 = 0 + (0) t + \frac{1}{2}(-9.8)t^2
```
```
4 = -4.9t^2
```
```
t^2 = -0.816
```
```
t = -0.90
```
We can now solve for the final velocity of the puck using the following equation:
```
v_y = -9.8t
```
Plugging in the known value for $t$, we get:
```
v_y = -9.8(-0.90)
```
```
v_y = 8.82 m/s
```
The final velocity of the puck in the vertical direction is 8.82 m/s downward.
* **Step 4: Find the angle.**
The angle of the velocity of the puck can be found using the following equation:
```
\theta = \tan^{-1} \left ( \frac{v_y}{v_x} \right )
```
Plugging in the known values for $v_y$ and $v_x$, we get:
```
\theta = \tan^{-1} \left ( \frac{8.82}{28.0} \right )
```
```
\theta = 12.6^\circ
```
Therefore, the direction of the velocity of the puck just before it hits the ground is 12.6 degrees below the horizontal.