\[m \frac{dv_x}{dt}= -kv_x\]
Separate variables to give
\[\frac{dv_x}{v_x} = - \frac{k}{m} dt\]
and integrate to give \[ln \frac{v_x}{v^0_x} = -\frac{kt}{m} \rightarrow v_x = v^0_x e^{-\frac{kt}{m}}\]
Applying Newton's Second Law vertically gives
\[m \frac{dv_y}{dt}= -kv_y - mg\]
Separating variables gives
\[m \frac{d v_y}{mg+kv_y} = - dt\]
Integration, with
\[v_y =v^0_y\]
at \[t=0\]
gives \[\frac{m}{k} ln(\frac{mg/k + v_y}{mg/k+v^0_y}) =-t \]
Rearranging this expression for
\[v_y\]
gives \[v_y = (\frac{mg}{k} +v^0_y) e^{- \frac{kt}{m}} - \frac{mg}{k}\]
At any point, the angle the velocity makes with the horizontal is
\[\theta\]
where \[tan \theta = \frac{v_y}{v_x} = \frac{(\frac{mg}{k} +v^0_y) e^{- \frac{kt}{m}} - \frac{mg}{k}}{v^0_x e^{-\frac{kt}{m}}}\]