Are this equations correct, in order to calculate the parabolic motion of an arrow with the computation of the drag with the air?

$$ \begin{cases} x(t)=\left(v_0-\frac{1/2C_DA\rho v_0^2}{m}t\right)\cos(\theta)t\\ y(t)=\left(v_0-\frac{1/2C_DA\rho v_0^2}{m}t\right)\sin(\theta)t-\frac{1}{2}gt^2+h \end{cases} $$

Update: correction.

$$ \vec{r}= \begin{vmatrix} \left(v_0-\frac{C_DA\rho v_0^2}{4m}t\right)\cos(\theta)t \\ \left(v_0-\frac{C_DA\rho v_0^2}{4m}t\right)\sin(\theta)t-\frac{1}{2} g t^2+h \\ 0 \end{vmatrix} $$

$$ \vec{a} = \ddot{\vec{r}} = \begin{vmatrix} -\frac{C_DA\rho v_0^2}{2m}\cos(\theta) \\ -\frac{C_DA\rho v_0^2}{2m}\sin(\theta) -g \end{vmatrix} $$

Update for AlanSE:

\begin{equation} \begin{split} m\frac{d^2 x(t)}{d t^2}&=-\frac{C_DA\rho}{2}\sqrt{\left(\frac{dx(t)}{dt}\right)^2+\left(\frac{dy(t)}{dt}\right)^2}\frac{dx(t)}{dt},\\ m\frac{d^2 y(t)}{dt^2}&=-mg-\frac{C_DA\rho}{2}\sqrt{\left(\frac{dx(t)}{dt}\right)^2+\left(\frac{dy(t)}{dt}\right)^2}\frac{d y(t)}{dt}. \end{split} \end{equation}

Related Cauchy problem:

\begin{equation} \begin{cases} \displaystyle x(0)=0,\\ \displaystyle y(0)=h. \end{cases} \end{equation}

  • $\begingroup$ It depends. Is the air friction term linear or quadratic? Also, how did you get those? In which approximation do you need them? Do you have to track an actual arrow, or is it for a homework? $\endgroup$ – Ferdinando Randisi Oct 28 '12 at 14:49
  • 2
    $\begingroup$ First things first... You shouldn't ask users to check for your home-made set of equations :-) $\endgroup$ – Waffle's Crazy Peanut Oct 28 '12 at 14:53
  • $\begingroup$ Get rid of that $v_0^2$ in the drag terms, you don't need it. The rest of terms preceding the square root is what I had termed $k$. THe equations you now have are basically the ones in my answer to your question, replacing $v_x$ and $v_y$ with $dx/dt$ and $dy/dt$. $\endgroup$ – Jaime Nov 2 '12 at 3:07
  • $\begingroup$ @Jaime Corrected. Sure, I have rewritten it only to be sure to have understood correctly what you have explained me... $\endgroup$ – Aurelius Nov 2 '12 at 14:33

Ok lets see. Differentiate the positions two times to arrive at the acceleration vector and see if it obeys Newtons Laws.

$$ \vec{r} = \begin{vmatrix} \left(v_0-\frac{C_DA\rho v_0^2}{2 m}t\right)\cos(\theta)t \\ \left(v_0-\frac{C_DA\rho v_0^2}{2 m}t\right)\sin(\theta)t-\frac{1}{2} g t^2+h \\ 0 \end{vmatrix} $$

$$ \vec{a} = \ddot{\vec{r}} = \begin{vmatrix} -\frac{C_DA\rho v_0^2}{m}\cos(\theta) \\ -\frac{C_DA\rho v_0^2}{m}\sin(\theta) -g \end{vmatrix} $$

So 1) the acceleration does not depend on the instantaneous velocity, only the initial velocity. 2) The drag force is missing the $\frac{1}{2}$ coefficient.

So the answer is no.

  • $\begingroup$ Please, look at my updated question, with your corrections, have I understood? And why you have said "the acceleration does not depend on the instantaneous velocity, only the initial velocity." I don't see the link. $\endgroup$ – Aurelius Nov 1 '12 at 13:11
  • $\begingroup$ @FormlessCloud, only $v_0$ is used in $\vec{a}$, not $v(t)$. Drag force changes with time as speed changes. $\endgroup$ – ja72 Nov 1 '12 at 18:37

As ja72 points out, the formulas you have produced are, apart from a missing $\frac{1}{2}$ factor, what you would get if drag was proportional to the square of the initial velocity, not the instantaneous one.

For quadratic drag, you need to solve the following pair of equations:

$$m \dot{v_x} = -k\sqrt{v_x^2 + v_y^2} v_x,$$ $$m \dot{v_y} = -mg-k\sqrt{v_x^2 + v_y^2} v_y.$$

As fas as I know, unless $v_x=0$, or $g=0$, the above pair of differential equations cannot be solved analitically in terms of elementary functions. So your formulas are not correct, mostly because there are no such formulas.

  • $\begingroup$ Could you explain the expressions that you have written? for example what is $k$, $v_x$ and $v_y$? $\endgroup$ – Aurelius Nov 1 '12 at 9:48
  • $\begingroup$ @FormlessCloud The $v_x$ and $v_y$ are the velocities, which are the derivative of your $x(t)$ and $y(t)$. To be a complete set of differential equations we need to write $dx/dt=v_x$ etc. You also should formally write your initial conditions in order to have a mathematically complete problem. $\endgroup$ – Alan Rominger Nov 1 '12 at 13:20
  • $\begingroup$ @AlanSE Look at my updated question, have I understood correctly? What is the name of the equations suggested from Jaime? $\endgroup$ – Aurelius Nov 1 '12 at 21:41

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