Parabolic motion and air drag

Question

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}

Answer 1

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.

Answer 2

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.