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 $$y''(t)=-h-ky'(t)\tag{5.1}$$ $$dy'=(-g-ky')dt\tag{5.2}$$ $$\frac{dy'}{-g-ky'}=dt\tag{5.3}$$ $$\frac{1}{k}\ln(g+ky')=-t+C\tag{5.4}$$


  1. Nina Henelsmith, Projectile Motion:Finding the Optimal Launch Angle, Whitman College, 2016; p. 24.
  • 2
    $\begingroup$ 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. $\endgroup$
    – David Z
    Commented Mar 28, 2020 at 5:51

1 Answer 1


$b$ 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 $\vec{\bf v}=x^{\prime}(t)\hat{\bf x}+y^{\prime}(t)\hat{\bf y}$.


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