1
$\begingroup$

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.
$\endgroup$
  • 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 Mar 28 at 5:51
2
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.