The initial velocity is purely horizontal, $\vec{v}_i = (v_{ix}, 0)$. The net force is gravity, $\vec{F}_{net} = (0, -mg)$. According to the impulse-momentum theorem, $\vec{F}_{net}t = m(\vec{v}_f - \vec{v}_i)$. In components:

$x: 0 \cdot t = m(v_{fx} - v_{ix}) \implies v_{fx} = v_{ix} = 10$ m/s $y: -mgt = m(v_{fy} - 0) \implies v_{fy} = -gt$

The final speed is the magnitude of the final velocity vector, $|\vec{v}_f| = \sqrt{v_{fx}^2 + v_{fy}^2}$.

$$|\vec{v}_f| = \sqrt{v_{ix}^2 + (-gt)^2} = \sqrt{(10 \text{ m/s})^2 + (-9.8 \text{ m/s}^2 \cdot 2 \text{ s})^2}$$