1
$\begingroup$

Here is the equation I obtained, representing the horizontal range of a projectile in air resistance as a function of its initial release angle-

$$x=\frac{V_0\cos(\theta)}{B}-\frac{V_0\cos(\theta)}{B}\,\exp\left(\frac{BV_0\sin(\theta)}{g}\right)\exp\left(W\left(\frac{BV_0\sin(\theta)\exp\left(\frac{BV_0\sin(\theta)}{-g}\right)}{-g}\right)\right), $$

where:

$x=$ horizontal distance travelled by projectile

$V_0=$ initial velocity of projectile

$g=$ acceleration due to gravity

$\theta=$ initial release angle

I have assumed that the force of air resistance acting on the projectile is equal to a constant k multiplied by the velocity of the projectile.

So, $F=kv.$

In my equation, $B= k/m,$ where $m$ is the mass of the projectile.

Also, $W(\cdot)$ is the $W$ Lambert function.

I have been trying to differentiate the expression with respect to $\theta$ in order find an expression for the value of $\theta$ that would result in maximum horizontal range, assuming all other terms in the equation are constant.

How can this be done?

$\endgroup$
  • 1
    $\begingroup$ According to this diagram, $X=0$. Can you show how you found the equation for $X$? $\endgroup$ – Peter Foreman Jul 8 at 14:18
  • 1
    $\begingroup$ Since the equation for $X$ is not too long, can you type it up? $\endgroup$ – automaticallyGenerated Jul 8 at 14:21
  • $\begingroup$ Curious about $F=kv.$ Is that the magnitude of the force equal to a constant times the magnitude of the velocity? $\endgroup$ – Adrian Keister Jul 8 at 14:41
  • $\begingroup$ Yes. k is a constant, which depends on the properties of the fluid that the projectile is travelling through. F is the force of air resistance, the magnitude of which is equal to the constant k times the magnitude of the velocity of the projectile. $\endgroup$ – LZR789 Jul 8 at 14:45
  • $\begingroup$ So what's the vector equation? I'm guessing $\mathbf{F}=-k\mathbf{v};$ is that correct? $\endgroup$ – Adrian Keister Jul 8 at 15:07
0
$\begingroup$

Starting from $\mathbf{F}=-k\,\mathbf{v}-mg\,\hat{\mathbf{j}}=m\dot{\mathbf{v}},$ and initial conditions $\mathbf{v}_0=v_0\langle\cos(\theta),\sin(\theta)\rangle$ and $\mathbf{x}_0=\langle 0,0\rangle,$ we obtain \begin{align*} m\ddot{x}&=-k\dot{x} \\ \ddot{x}&=-B\dot{x} \\ \dot{x}&=Ce^{-Bt} \\ \dot{x}&=v_0\cos(\theta)\,e^{-Bt} \\ x&=-\frac{v_0\cos(\theta)}{B}\,e^{-Bt}+C\\ x(t)&=\frac{v_0\cos(\theta)}{B}\,\left(1-e^{-Bt}\right),\;\text{and} \\ m\ddot{y}&=-k\dot{y}-mg \\ \ddot{y}&=-B\dot{y}-g \\ \int\frac{d \dot{y}}{B\dot{y}+g}&=-\int dt\\ \frac1B\int\frac{d \dot{y}}{\dot{y}+g/B}&=-t+C\\ \frac1B\,\ln|\dot{y}+g/B|&=-t+C\\ \ln|\dot{y}+g/B|&=-Bt+C, \quad\text{absorbing into constant}\\ \dot{y}+g/B&=Ce^{-Bt},\quad\text{switching to multiplicative constant}\\ v_0\sin(\theta)+\frac{g}{B}&=C\\ \dot{y}&=\left(v_0\sin(\theta)+\frac{g}{B}\right)e^{-Bt}-\frac{g}{B}\\ y&=-\left(\frac{v_0\sin(\theta)+\frac{g}{B}}{B}\right)e^{-Bt}-\frac{gt}{B}+C\\ y(t)&=\frac{v_0\sin(\theta)+mg/k}{B}\left(1-e^{-Bt}\right)-\frac{mgt}{k}. \end{align*} Solving the second equation $y(t)=0$ for $t$ yields $$t_{\text{range}}=\frac{mg W\left(-\frac{e^{-\frac{k v_0 \sin (\theta )}{mg}-1} (mg+k v_0 \sin (\theta ))}{mg}\right)+mg+k v_0 \sin (\theta )}{B mg}. $$ Plugging this back into $x$ yields $$x_{\text{range}}(\theta)=\frac{v_0 \cos (\theta )}{B} \left(1-\exp \left(- W\left(-e^{-\frac{k v_0 \sin (\theta )}{mg}-1} (1+(k v_0/(mg)) \sin (\theta ))\right)+1+(k v_0/(mg)) \sin (\theta )\right)\right). $$ Now it is a fact that $$\dot{W}(t)=\frac{W(t)}{t(1+W(t))}. $$ Using this, we can differentiate this expression (it's quite a monster) to obtain $$x_{\text{range}}'(\theta)=-\frac{v_0 \left(mg W\left(-\frac{e^{-\frac{k v_0 \sin (\theta )}{mg}-1} (mg+k v_0 \sin (\theta ))}{mg}\right)+mg+k v_0 \sin (\theta )\right) \left((mg \sin (\theta )+k v_0) W\left(-\frac{e^{-\frac{k v_0 \sin (\theta )}{mg}-1} (mg+k v_0 \sin (\theta ))}{mg}\right)+\sin (\theta ) (mg+k v_0 \sin (\theta ))\right)}{B (mg+k v_0 \sin (\theta ))^2 \left(W\left(-\frac{e^{-\frac{k v_0 \sin (\theta )}{mg}-1} (mg+k v_0 \sin (\theta ))}{mg}\right)+1\right)}. $$ Solving $x_{\text{range}}'(\theta)=0$ for $\theta$ will be tricky. We can eliminate the denominator and just solve $$-v_0 \left(mg W\left(-\frac{e^{-\frac{k v_0 \sin (\theta )}{mg}-1} (mg+k v_0 \sin (\theta ))}{mg}\right)+mg+k v_0 \sin (\theta )\right) \left((mg \sin (\theta )+k v_0) W\left(-\frac{e^{-\frac{k v_0 \sin (\theta )}{mg}-1} (mg+k v_0 \sin (\theta ))}{mg}\right)+\sin (\theta ) (mg+k v_0 \sin (\theta ))\right) =0.$$ Mathematica does not immediately find a solution, but we note that this expression is factored. That is, we can solve either of \begin{align*} mg W\left(-\frac{e^{-\frac{k v_0 \sin (\theta )}{mg}-1} (mg+k v_0 \sin (\theta ))}{mg}\right)+mg+k v_0 \sin (\theta )&=0,\quad\text{or}\\ (mg \sin (\theta )+k v_0) W\left(-\frac{e^{-\frac{k v_0 \sin (\theta )}{mg}-1} (mg+k v_0 \sin (\theta ))}{mg}\right)+\sin (\theta ) (mg+k v_0 \sin (\theta ))&=0. \end{align*} The first one initially looks a bit more promising. Let $s=\sin(\theta),$ so that we have \begin{align*} mg W\left(-\frac{e^{-\frac{k v_0 s}{mg}-1} (mg+k v_0 s)}{mg}\right)+mg+k v_0 s&=0 \\ W\left(-\frac{e^{-\frac{k v_0 s}{mg}-1} (mg+k v_0 s)}{mg}\right)+1+(k v_0/(mg)) s&=0 \\ W\left(-\frac{e^{-\frac{k v_0 s}{mg}-1} (mg+k v_0 s)}{mg}\right)&=-1-\frac{kv_0s}{mg} \\ -\frac{e^{-\frac{k v_0 s}{mg}-1} (mg+k v_0 s)}{mg}&=-\left(\frac{mg+kv_0s}{mg}\right)\exp\left(-1-\frac{kv_0s}{mg}\right), \end{align*} which you can see is a tautology! That is, it conveys no information. Therefore, any solution must be from the other factor. Let us work on that (again, $s=\sin(\theta)$) \begin{align*} (mgs+k v_0) W\left(-\frac{e^{-\frac{k v_0 s}{mg}-1} (mg+k v_0 s)}{mg}\right)+s (mg+k v_0 s)&=0 \\ (mg s+k v_0) W\left(-\frac{e^{-\frac{k v_0 s}{mg}-1} (mg+k v_0 s)}{mg}\right)&=-s (mg+k v_0 s) \\ W\left(-\frac{e^{-\frac{k v_0 s}{mg}-1} (mg+k v_0 s)}{mg}\right)&=-s\,\frac{mg+kv_0s}{mgs+kv_0} \\ -\frac{e^{-\frac{k v_0 s-mg}{mg}} (mg+k v_0 s)}{mg}&=-s\,\frac{mg+kv_0s}{mgs+kv_0}\,\exp\left(-s\,\frac{mg+kv_0s}{mgs+kv_0}\right) \\ \exp\left(-\frac{k v_0 s-mg}{mg}\right) &=\frac{mgs}{mgs+kv_0}\,\exp\left(-s\,\frac{mg+kv_0s}{mgs+kv_0}\right). \end{align*} In canceling $mg+kv_0s,$ we are eliminating the solution $s=-mg/(kv_0),$ corresponding to $$\theta=\arcsin\left(-\frac{mg}{kv_0}\right). $$ A negative answer seems physically impossible, so we rule it out. We continue: \begin{align*} \exp\left(\frac{mg-k v_0 s}{mg}\right) &=\frac{mgs}{mgs+kv_0}\,\exp\left(-s\,\frac{mg+kv_0s}{mgs+kv_0}\right)\\ \frac{mg-k v_0 s}{mg}&=\ln\left(\frac{mgs}{mgs+kv_0}\right)-s\,\frac{mg+kv_0s}{mgs+kv_0}\\ \frac{g-B v_0 s}{g}&=\ln\left(\frac{gs}{gs+Bv_0}\right)-s\,\frac{g+Bv_0s}{gs+Bv_0}. \end{align*} Unfortunately, this is pretty much as far as we can go. $s$ appears in far too many places in this equation to think that there is an analytical solution. My recommendation is to take this equation, plug in the values you need for $m, g, k,$ and $v_0,$ and then solve it numerically.

$\endgroup$
  • $\begingroup$ Thanks for your help. I got the same expression as you for x(t) but arrived at something slightly different for y(t) using differential equations. I started from: (d^2 y/dt^2)+B(dy/dt)=-g and got y= -gt/b + (v sinθ/B)(1- e^-Bt). How did you end up with k in your expression? $\endgroup$ – LZR789 Jul 9 at 12:46
  • $\begingroup$ I think you lost an $m$ somewhere: \begin{align*}F_y&=-kv_y-g \\ m\ddot{y}&=-k\dot{y}-g \\ \ddot{y}&=-\frac{k}{m}\,\dot{y}-\frac{g}{m} \\ \ddot{y}+B\dot{y}&=-\frac{g}{m}. \end{align*} The $k$ is bound up in $B,$ and depending on how you manipulate things, you might get something slightly different. $\endgroup$ – Adrian Keister Jul 9 at 14:46
  • $\begingroup$ Surely F= -kv -mg as g is the acceleration due to gravity so -mg would be the force experienced due to gravity. This would result in a = -bv-g so (d^2 y/dt^2)+B(dy/dt)= -g. $\endgroup$ – LZR789 Jul 9 at 15:44
  • $\begingroup$ Oh, you're right, my mistake. I'll need to go through and replace all $g$'s with $mg$'s to fix it. $\endgroup$ – Adrian Keister Jul 9 at 15:45
  • $\begingroup$ Ok, I think I've fixed the issue. $\endgroup$ – Adrian Keister Jul 9 at 18:53

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.