# 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. – David Z Mar 28 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 $$\vec{\bf v}=x^{\prime}(t)\hat{\bf x}+y^{\prime}(t)\hat{\bf y}$$.