Projectile with linear drag [closed]

Question

A particle with mass $m$ travels with speed $v_0$ and angle $\theta$ compared to the horizontal ground. Air resistance given as $F_R = -m \alpha v$, where $v$ is the particle's speed, is working against the projectile.

The particle's motion is described as

$$\frac{{dv_x} }{dt} + \alpha v_x = 0$$

$$\frac{{dv_y} }{dt} + \alpha v_y = -g$$

where

$$v_x = \frac{{dx} }{dt}$$

$$v_y = \frac{{dy} }{dt}$$

Show that the particles position is given as

$$ x(t) = \frac{{1} }{\alpha}(v_0 \cos \theta)(1-e^{-\alpha t})$$ $$ y(t) = \frac{{1} }{\alpha}(v_0 \sin \theta + \frac{g}{\alpha})(1-e^{-\alpha t}) - \frac{g}{\alpha}t$$

I understand how to set up the equations and do the integrals (at least a part of it) and get the following result:

$$ x(t) = \frac{{1} }{\alpha}(1-e^{-\alpha t})$$

What I don't understand is where the $v_0 cos \theta$ is coming from. I'm thinking it has something to do with decomposing forces in $x$ and $y$ direction, but I can't seem to work it out and where it goes in the equation.

(I haven't started with $y(t)$ yet).

Answer 1

I think, you forgot to add the integration constant. Integration of ${dv_x \over dt}+\alpha v_x=0$ gives $ln {v_x}=\alpha t+c_0$. After some algebraic manipulation we can write $v_x=Ae^{\alpha t}$, where $A=e^{c_0}$, and applying the initial condition gives $A=v_{x0}$ and from component decomposition we know $v_{x0}=v_0 cos \theta $. So now we're left with the following differential $eq^n$-$${dx \over dt}=v_0 cos \theta e^{ \alpha t}$$ whose $sol^n$ is $x(t)={1 \over \alpha} v_0 cos \theta e^{\alpha t}+c_1$. Applying the initial condition $x=0$ at $t=0$ , we get $c_1= -{1 \over \alpha}v_0 cos \theta$. Substituting it in the $x(t)$ expression you'll get your desired expression for $x(t)$.