# Defining variable in projectile motion

I'm trying to understand drag on projectile motion but I don't know what this variable $$b$$ in Eq.1 below on p. 24 in this document is.

We define the drag force to be $$\mathbf{F}_D$$ and the gravitational force is $$\mathbf{F_g}$$. We have $$m\mathbf{a}=\mathbf{F}=\mathbf{F_g}+\mathbf{F_D}=mg\mathbf{\hat y}-b(\mathbf{\hat x}+\mathbf{\hat y}),\tag{1}$$ and letting $$k=b/m$$, we can separate the above equation into $$x-$$ and $$y-$$equations. We have $$x''(t)=-kx'(t), \quad y''(t)=-h-ky'(t).\tag{2}$$ Next we will solve the above differential equations using the initial conditions: $$x(0)=0; \quad y(0)=h;\tag{3.1}$$ $$x'(0)=v\cos\theta; \quad y'(0)=v\sin\theta;\tag{3.2}$$ where $$v$$ is the initial velocity of the projectile. Using separation of variables to solve the $$x-$$equation, we obtain $$x''(t)=-kx'(t),\tag{4.1}$$ $$x'(t)=Ce^{-kt}=v\cos\theta e^{-kt}\tag{4.2}$$ $$x(t)=-\frac{v\cos\theta}{k}e^{-kt}+C=\frac{v\cos\theta}{k}\left(1-e^{-kt}\right).\tag{4.3}$$ Similarly, for the $$y-$$equation, we have $$y''(t)=-h-ky'(t)\tag{5.1}$$ $$dy'=(-g-ky')dt\tag{5.2}$$ $$\frac{dy'}{-g-ky'}=dt\tag{5.3}$$ $$\frac{1}{k}\ln(g+ky')=-t+C\tag{5.4}$$

References:

1. Nina Henelsmith, Projectile Motion:Finding the Optimal Launch Angle, Whitman College, 2016; p. 24.
• Hi WaterRocket123, it's against our rules to post images of text you want to quote. Please type it out instead so it can be indexed by search engines. For formulas, use MathJax. Commented Mar 28, 2020 at 5:51

$$b$$ is the drag constant for a linear drag force. OP's question might have been spurred by the fact that the first eq. on p. 24 has a typo: The factor $$\hat{\bf x}+\hat{\bf y}$$ should have been the velocity $$\vec{\bf v}=x^{\prime}(t)\hat{\bf x}+y^{\prime}(t)\hat{\bf y}$$.