Answer
The answer is:
$$\vec{v}_f=(2\text{ m/s })\hat{x}+(-4\text{ m/s })\hat{y}.$$
Work Step by Step
The horizontal component (along $\hat{x}$ direction) is constant while the vertical component (along $\hat{y}$ direction) changes with motion because the gravitational acceleration points downwards vertically, and from initial $v_{0y}$ it goes to $0$ at the highest point and then increases in magnitude, but starts pointing downwards, until it achieves $-v_{0y}$ just before the landing. This means that the final velocity, just before the landing is
$$\vec{v}_f=(2\text{ m/s })\hat{x}+(-4\text{ m/s })\hat{y}.$$
