Air Friction Stops my Projectile

As per the given question,

$\begin{aligned} \vec{F_{net}} &= -b (v_x \hat i + v_y \hat j) - mg \hat j \\ F_x &= -b (v_x ) \\ ma_x &= -b (v_x ) \\ m v_x \dfrac{dv_x}{dx} &= -b (v_x ) \\ dv_x &= \dfrac{-b}{m}\times dx \end{aligned}$

Integrating for horizontal direction of velocity,

$\begin{aligned} \displaystyle{\int^{0}_{v_{0} \cos 60} d|\vec{v_{x}}|} &= \displaystyle{\int^{x}_{0} \dfrac{-b}{m}\times dx} \\ (0-\dfrac{v_{0}}{2}) &= \dfrac{-b}{m} x \\ \end{aligned} \\ \boxed{x = \dfrac{mv_{0}}{2b}} \\ \alpha = 1,\, \beta=2 \\ \alpha+\beta = \boxed{3}$

As we only need horizontal information, we'll use Newton's second law of motion: $m\ddot{x}=-b\dot{x}$ $\Rightarrow D(D+\frac{b}{m})(x(t))=0$ Also we have the initial conditions $x(0)=0$ and $\dot{x}(0)=v_{0}\cdot \cos(\frac{\pi}{3})=\frac{v_{0}}{2}$ the differential equation gives: $x(t)=\frac{mv_{0}}{2b}(1-e^{-\frac{b}{m}t})$ So the answer is $1+2=3$