# Projectile with linear drag [closed]

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).

• What I don't understand is where the $v_0\cos\theta$ is coming from. From the initial conditions. – Gert Aug 29 at 14:27
• $v_{x,0}=v_0\cos\theta$, $v_{y,0}=v_0\sin\theta$ – Gert Aug 29 at 15:50

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 $$x(t)$$.