The only force on the bullet is gravity, $\vec{F}_{net} = -mg\hat{j}$. The bullet starts and ends at the same height. Due to symmetry, the final vertical velocity is the negative of the initial vertical velocity: $v_{fy} = -v_{iy} = -v_0\sin\theta$. Applying the impulse-momentum theorem to the vertical motion:

$$F_y t = \Delta p_y = m(v_{fy} - v_{iy})$$ $$-mgt = m(-v_0\sin\theta - v_0\sin\theta) = -2mv_0\sin\theta$$ $$t = \frac{2v_0\sin\theta}{g} = \frac{2(100 \text{ m/s})\sin(60^\circ)}{9.8 \text{ m/s}^2}$$

For the vector diagram, the initial momentum $\vec{p}_i$ points up and to the right. The final momentum $\vec{p}_f$ points down and to the right with the same horizontal component. The change in momentum $\Delta\vec{p} = \vec{p}_f - \vec{p}_i$ is a vector pointing vertically downward. The impulse $\vec{I}_{net} = \vec{F}_{net}t$ also points vertically downward, consistent with $\vec{I}_{net} = \Delta\vec{p}$.