2D Motion of Projectile in a Fluid with Friction

A 2D projectile with mass \(m\) is moving in a uniform gravitational field \(\overrightarrow{g}\). The projectile is launched at an initial velocity \({\overrightarrow{v}}_0\) from a reference point \(\left(x_0,y_0\right)\) an angle \({\varphi }_0\). The drag force \({\overrightarrow{f}}_D\) is assumed to be proportional to the speed \(\overrightarrow{v}\) raised to the power \(n\).

Figure 1

From Newton’s second law and Figure 1, we can write the following vector-equation:

\[\sum{\overrightarrow{F}}=-mg\ \overrightarrow{j}+{\overrightarrow{f}}_D=m\ \overrightarrow{a} \label{1}\]

The expression of the projectile’s speed \(\overrightarrow{v}\) is given by:

\[\begin{aligned} &{\overrightarrow{v}=\overrightarrow{v}_x}+{\overrightarrow{v}_y}=v_x\ \overrightarrow{i}+v_y\ \overrightarrow{j}  \\[4pt]
 &{\overrightarrow{u}}_v=\dfrac{\overrightarrow{v}}{\left\|\overrightarrow{v}\right\|}=\dfrac{\overrightarrow{v}}{v}=\dfrac{v_x\ \overrightarrow{i}+v_y\ \overrightarrow{j}}{\sqrt{v^2_x+v^2_y}}=\left\langle {\mathrm{cos} \varphi \ }\ ,{\sin \varphi \ }\right\rangle ={\overrightarrow{u}_t}  \end{aligned} \label{2}\]

The expression of the drag force \({\overrightarrow{f}}_D\) is given by:

\[\overrightarrow{f}_D = -C_Dv^n \overrightarrow{u}_t \label{3}\]

where \(C_D\) is the coefficient of drag.

With Equation \ref{2} and Equation \ref{3} into Equation \ref{1}, we get the 2D vector- Equation of the motion in Cartesian coordinates:

\[-mg\ \overrightarrow{j}-C_Dv^n\ \left(\frac{v_x\ \overrightarrow{i}+v_y\ \overrightarrow{j}}{v}\right)=m\ \left({\overrightarrow{a}}_x\ \overrightarrow{i}+{\overrightarrow{a}}_y\overrightarrow{j}\right) \nonumber\]

After regrouping the terms, we obtain the following differential Equation system:

\[\left\{\begin{array}{l}
\vec{\imath} \rightarrow-D v^{n-1} v_x=\frac{d v_x}{d t} \\
\vec{\jmath} \rightarrow-g-D v^{n-1} v_y=\frac{d v_y}{d t}
\end{array}\right. \label{4} \]

where \(D=\frac{C_D}{m}\) is constant.

Generally, the system Equation \ref{4} is fully algebraically solvable for \(n=1\), otherwise we have to proceed with numerical iterations or graphic methods to find the solutions.

1: D=0

Is the case of the motion in the vacuum and Equation \ref{4} is reduced to:

\[\left\{\begin{array}{l}
0=\frac{d v_x}{d t} \\
-g=\frac{d v_y}{d t}
\end{array}\right. \label{5}\]

It is the classic parametric Equation of an inverted parabola, by consecutive integration from \(\left[0,t\right]\) of Equation \ref{5} we get:

\[\left\{\begin{array} { r l } 
{ v _ { x } } & { = v _ { x _ { 0 } } } \\
{ - g t } & { = v _ { y } - v _ { y _ { 0 } } }
\end{array} \Rightarrow \left\{\begin{array} { c } 
{ x - x _ { 0 } = v _ { x _ { 0 } } t } \\
{ \int _ { 0 } ^ { t } ( v _ { y _ { 0 } } - g t ) d t = \int _ { y _ { 0 } } ^ { y } \frac { d y } { d t } d t }
\end{array} \rightarrow \left\{\begin{array}{l}
x-x_0=v_{x_0} t \\
y-y_0+t\left(v_{y_0}-\frac{g t}{2}\right)
\end{array}\right.\right.\right. \label{6}\]

Eliminating the parameter \(t\) we get:

\[\left\{\begin{array}{l}
\frac{x-x_0}{v_{x_0}}=t \\
y-y_0=v_{y_0} \frac{x-x_0}{v_{x_0}}-\frac{g}{2}\left(\frac{x-x_0}{v_{x_0}}\right)^2
\end{array}\right. \label{7}\]

Taking into account Equation \ref{2}, after the due simplifications, Equation \ref{7}.2 becomes:

\[y\left(x-x_0\right)=y_0+\frac{x-x_0}{\cos \varphi_0}\left(\sin \varphi_0-\frac{g}{2 v_0^2} \frac{x-x_0}{\cos \varphi_0}\right) \label{8}\]

Figure 2

2: D≠0

2.1: n=1

Is the case of the motion in the dense medium (liquid) and the set Equation \ref{4} becomes:

\[\left\{\begin{array}{l}
\vec{\imath} \rightarrow-D v_x=\frac{d v_x}{d t} \\
\vec{\jmath} \rightarrow-g-D v_y=\frac{d v_y}{d t}
\end{array}\right. \label{9}\]

We can express the constant \(D\) in terms of “Terminal Velocity” \(v_T\) and \(g\), from Equation \ref{4}.2 we can write:

\[-C_D\left(v_y=v_T\right)-mg=0\ \Rightarrow \ D=\frac{g}{v_T} \label{10}\]

With Equation \ref{10} into Equation \ref{9}, we have the equivalent set of D.E. of the motion:

\[\left\{\begin{array} { l l } 
{ - D v _ { x } = \frac { d v _ { x } } { d t } }  \\
{ - g - D v _ { y } = \frac { d v _ { y } } { d t } } 
\end{array} \Leftrightarrow \left\{\begin{array}{l}
-g \frac{v_x}{v_T}=\frac{d v_x}{d t} \\
-g\left(1+\frac{v_y}{v_T}\right)=\frac{d v_y}{d t}
\end{array}\right.\right. \label{11}\]

2.1.1: The Velocities

By integration from \(\left[0,t\right]\) of Equation \ref{9} or Equation \ref{11}, we get:

\[\left\{\begin{array}{l}
-\frac{g}{v_T} \int_0^t d t=\int_{v_{x_0}}^{v_x} \frac{d v_x}{v_x} \Rightarrow-\frac{g}{v_T} t=\log \frac{v_x}{v_{x_0}} \\
-g \int_0^t d t=\int_{v_{y_0}}^{v_y} \frac{d v_y}{1+\frac{v_y}{v_T}} \Rightarrow-\frac{g}{v_T} t=\log \left(\frac{v_T+v_y}{v_T+v_{y_0}}\right)
\end{array}\right. \nonumber\]

\[\left\{\begin{array}{l}
v_x(t)=v_{x_0} e^{-\frac{g}{v_T} t}=v_0 \cos \varphi_0 e^{-D t} \\
v_y(t)=\left(v_T+v_{y_0}\right) e^{-\frac{g}{v_T} t}-v_T=\left(\frac{g}{D}+v_0 \sin \varphi_0\right) e^{-D t}-\frac{g}{D}
\end{array}\right. \label{12}\]

Combining Equations \ref{2.2} and \ref{1.2}, we obtain:

\[v=\sqrt{{\left({v_0\cos {\varphi }_0\ }e^{-D\ t}\ \right)}^2+{\left[\left(\frac{g}{D}+v_0{\sin {\varphi }_0\ }\right){\ e}^{-D\ t}-\frac{g}{D}\right]}^2} \label{13}\]

Figure 3

Note

1. Rewriting Equation \ref{12}.2 as follows: \[v_y\left(t\right)=\left(\frac{g}{D}+v_0{\sin {\varphi }_0\ }\right){\ e}^{-D\ t}-\frac{g}{D}=v_0{\sin {\varphi }_0\ }{\ e}^{-D\ t}-\frac{g}{D}\left(1-{\ e}^{-D\ t}\right)\nonumber\] We find: \[{\lim_{D\to 0} v_y\left(t\right)\ }=v_0{\sin {\varphi }_0\ }-g{\lim_{D\to 0} \left[\frac{1-{\ e}^{-D\ t}}{D}\right]\ } \nonumber\] \[{\text{lim}_{D\to 0} \left[\frac{1-{\ e}^{-D\ t}}{D}\right]\ }={\lim_{D\to 0} \left[\frac{\frac{d}{dD}\left(1-{\ e}^{-D\ t}\right)}{\frac{d}{dD}\left(D\right)}\right]\ }=t\Rightarrow {\lim_{D\to 0} v_y\left(t\right)\ }=v_0{\sin {\varphi }_0\ }-gt \nonumber\] And we get back Equation \ref{6}.2.
2. From Equation \ref{13} and Figure 3 we can see that the Terminal Velocity \(v_T\) will be never reached: \[v=\sqrt{{\left(v_{x_0}e^{-\frac{g}{v_T}\ t}\ \right)}^2+{\left[\left(v_T+v_{y_0}\right){\ e}^{-\frac{g}{v_T}\ t}-v_T\right]}^2}=v_T \nonumber\] \[\Rightarrow \ \left\{ \begin{array}{c} v_x\left(t\right)=v_{x_0}e^{-\frac{g}{v_T}\ t}=0\ \Rightarrow \ t\to \infty \ \\ v_y\left(t\right)=\left(v_T+v_{y_0}\right){\ e}^{-\frac{g}{v_T}\ t}-v_T=v_T\Rightarrow \ t\to \infty \end{array} \right. \nonumber\]

2.1.2: The Coordinates (x,y)

By integration from \(\left[0,t\right]\) of Equation \ref{12}, we get:

\[\left\{ \begin{array}{c} x\left(t\right)=\int^x_{x_0}{\frac{dx\left(t\right)}{dt}}dt=\int^t_0{v_x\left(t\right)}dt=v_{x_0}\ \int^t_0{e^{-\frac{g}{v_T}\mathrm{\ }t}}dt\ \mathrm{\ }\ \\ y\left(t\right)=\int^y_{y_0}{\frac{dy\left(t\right)}{dt}}dt=\int^t_0{v_y\left(t\right)}dt=\ \int^t_0{\left[\left(v_T+v_{y_0}\right){\ e}^{-\frac{g}{v_T}\mathrm{\ }t}-v_T\right]}dt\ \end{array} \right. \nonumber\]

\[\left\{\begin{array}{l}
x(t)=x_0+\frac{v_{x_0} v_T}{g}\left(1-e^{-\frac{g}{v_T} t}\right)=x_0+\frac{v_0 \cos \varphi_0}{D}\left(1-e^{-D t}\right) \\
y(t)=\left\{\begin{array}{l}
y_0+v_T\left[\frac{v_{y_0}+v_T}{g}\left(1-e^{-\frac{g}{v_T} t}\right)-t\right] \\
y_0+\frac{g}{D}\left[\left(\frac{v_0 \sin \varphi_0}{g}+\frac{1}{D}\right)\left(1-e^{-D t}\right)-t\right]
\end{array}\right.
\end{array}\right. \label{14}\]

From Equation \ref{14}.1 we get:

\[t\left(x-x_0\right)=\frac{v_T}{g} \log \left[\frac{v_{x_0} v_T}{v_{x_0} v_T-g\left(x-x_0\right)}\right]=\frac{1}{D} \log \left[\frac{v_0 \cos \varphi_0}{v_0 \cos \varphi_0-D\left(x-x_0\right)}\right]\label{15}\]

With Equation \ref{15} into \ref{14}.2, we have the expression of \(y\left(x\right)\):

\[y\left(x-x_0\right)=\left\{\begin{array}{l}
y_0+v_T\left\{\frac{\left(v_{x_0}+v_T\right)\left(x-x_0\right)}{v_{x_0} v_T}-\frac{v_T}{g} \log \left[\frac{v_{x_0}}{v_{x_0}-\frac{g}{v_T}\left(x-x_0\right)}\right]\right\} \\
\left.y_0+\frac{g}{D}\left\{\left(\frac{v_0 \sin \varphi_0}{g}+\frac{1}{D}\right) \frac{D\left(x-x_0\right)}{v_0 \cos \varphi_0}-\frac{1}{D} \log \left[\frac{v_0 \cos \varphi_0}{v_0 \cos \varphi_0-D\left(x-x_0\right)}\right]\right\} \right)
\end{array}\right. \label{16}\]

Figure 4

Figure 5

Note

1. From Equation \ref{14}.1 we have: \[\lim _{D \rightarrow 0} x(t)=x_0+v_0 \cos \varphi_0 \lim _{D \rightarrow 0}\left(\frac{1-e^{-D t}}{D}\right)=x_0+v_0 \cos \varphi_0 t \nonumber\] And we find back Equation \ref{7}.1.
2. Rewriting Equation \ref{14}.2 as follows: \[y(t)=y_0+v_0 \sin \varphi_0\left(\frac{1-e^{-D t}}{D}\right)+\frac{g}{D}\left(\frac{1-e^{-D t}}{D}\right)-\frac{g}{D} \nonumber\] We find: \[\lim _{D \rightarrow 0} y(t)=y_0+\left\{\begin{array}{c} v_0 \sin \varphi_0 \lim _{D \rightarrow 0}\left(\frac{1-e^{-D t}}{D}\right)=v_0 \sin \varphi_0 t+ \\ g \lim _{D \rightarrow 0}\left(\frac{1-e^{-D t}-D t}{D^2}\right)=g \frac{0}{0} \end{array}\right. \nonumber\] \[{\lim_{D\to 0} \left[\frac{\frac{d}{dD}\left({1-e}^{-D\mathrm{\ }t}-Dt\right)}{\frac{d}{dD}\left(D^2\right)}\right]\ }={\lim_{D\to 0} \left[\frac{t\ \left(e^{-D\mathrm{\ }t}-1\right)}{2D}\right]\ }=\frac{t}{2}{\lim_{D\to 0} \left(\frac{e^{-D\mathrm{\ }t}-1}{D}\right)\ }=-\frac{t^2}{2} \nonumber\] Therefore: \[{\lim_{D\to 0} y\left(t\right)\ }=y_0+{v_0\mathrm{\ sin} {\varphi }_0\ }t-\frac{{gt}^2}{2} \nonumber \] And we get back Equation \ref{7}.2.

The range \(x_R\) is given by Equation \ref{16}.2 as:

\[y\left(x_R\right)=y_0+\frac{g}{D}\left\{\left(\frac{v_0 \sin \varphi_0}{g}+\frac{1}{D}\right) \frac{D x_R}{v_0 \cos \varphi_0}-\frac{1}{D} \log \left[\frac{v_0 \cos \varphi_0}{v_0 \cos \varphi_0-D x_R}\right]\right\}=0 \label{17}\]

Or resolving the system Equation \ref{14} in \(t_R\):

\[\left\{\begin{array}{l}
x_R\left(t_f\right)=\frac{v_{x_0} v_T}{g}\left(1-e^{-\frac{g}{v_T} t_f}\right)=\frac{v_0 \cos \varphi_0}{D}\left(1-e^{-D t_f}\right) \\
0=\left\{\begin{array}{l}
v_T\left[\frac{v_{y_0}+v_T}{g}\left(1-e^{-\frac{g}{v_T} t_f}\right)-t_f\right] \\
\frac{g}{D}\left[\left(\frac{v_0 \sin \varphi_0}{g}+\frac{1}{D}\right)\left(1-e^{-D t_f}\right)-t_f\right]
\end{array}\right.
\end{array}\right. \label{18}\]

Equations Equation \ref{17} or Equation \ref{18} can be only solved numerically or graphically.

2.2: n=2

Is the case of the motion in the gaseous medium and the set Equation \ref{4} becomes:

\[\left\{\begin{array}{l}
\vec{\imath} \rightarrow-D v v_x=\frac{d v_x}{d t} \\
\vec{\jmath} \rightarrow-g-D v v_y=\frac{d v_y}{d t}
\end{array}\right. \label{19}\]

Figure 6

2.2.1: Graph r(φ)

Solving for \(\overrightarrow{r}\left(\varphi \right)=\left\langle \ x\left(s\left(\varphi \right)\right)\ ,y\left(s\left(\varphi \right)\right)\ \right\rangle\) is not possible to integrate algebraically the set Equation \ref{19}, but it is possible with the combined help of the “intrinsic coordinates” to simplify the Equations. From “Curvilinear Motion” doc we have:

\[\left\{ \begin{array}{c} {\overrightarrow{u}}_T\to F_t=m\left(\frac{dv}{dt}=\frac{d^2s}{dt^2}\right) \\ {\overrightarrow{u}}_n\to \ F_n=m\left[\frac{v^2}{\rho }=\frac{1}{\rho }{\left(\frac{ds}{dt}\right)}^2\right] \end{array} \right. \tag{46} \label{46}\]

Note that from Figure 6, \(\frac{d\varphi }{dt}<0\), therefore: \[{\overrightarrow{u}}_T=\left\langle {\mathrm{cos} \varphi \ },{\sin \varphi \ }\right\rangle \Rightarrow {\overrightarrow{u}}_n=\frac{d{\overrightarrow{u}}_T}{d\varphi }=\left\langle -{\sin \varphi \ },{\mathrm{cos} \varphi \ }\right\rangle \nonumber\]

From Equation \ref{19}.1 combined with the intrinsic coordinates, we get in the \(\overrightarrow{i}\) direction:

\[-D\ v\mathrm{\ }v_x=\frac{dv_x}{dt}=-D\ \frac{d\mathrm{s}}{dt}v_x=\frac{dv_x}{dt}\Rightarrow -D\int^s_{s_i}{ds}=\int^{v_x}_{v_{x_i}}{\frac{dv_x}{v_x}} \nonumber\]

\[-D\left(s-s_i\right)=\log \left(\frac{v_x}{v_{x_i}}\right) \Rightarrow v_x=v_{x_i} e^{-D\left(s-s_i\right)}  \label{20}\]

From Equation \ref{20} and the definition of \(v\), we can write:

\[v=v_i \frac{\cos \varphi_i}{\cos \varphi} e^{-D\left(s-s_i\right)} \label{21}\]

From \ref{46}.2 combined with Equation \ref{21}, we get the \(\overrightarrow{n}\) direction:

\[-g \cos \varphi=\frac{v^2}{\rho}=\frac{d \varphi}{d s}\left(v_i \frac{\cos \varphi_i}{\cos \varphi} e^{-D\left(s-s_i\right)}\right)^2 \nonumber\]

\[\frac{g}{\left(v_i \cos \varphi_i\right)^2} \int_{s_i}^s e^{2 D\left(s-s_i\right)} d s=\frac{g}{\left(v_i \cos \varphi_i\right)^2} \int_0^{s-s_i} e^{2 D u} d u=-\int_{\varphi_i}^{\varphi} \frac{d \varphi}{(\cos \varphi)^3} \nonumber\]

Since:

\[\left\{\begin{aligned}
\int \frac{d \varphi}{(\cos \varphi)^n} & =\frac{1}{n-1}\left[\frac{\sin \varphi}{(\cos \varphi)^{n-1}}+(n-2) \int \frac{d \varphi}{(\cos \varphi)^{n-2}}\right]+C \\
\int \frac{d \varphi}{\cos \varphi} & =\log \left(\frac{1+\sin \varphi}{\cos \varphi}\right)+C
\end{aligned}\right. \nonumber\]

After integration, in the \(\overrightarrow{n}\) direction we have:

\[\frac{g\left(e^{2 D\left(s-s_i\right)}-1\right)}{D\left(v_i \cos \varphi_i\right)^2}=\frac{\sin \varphi_i}{\left(\cos \varphi_i\right)^2}-\frac{\sin \varphi}{(\cos \varphi)^2}+\log \left[\frac{\left(1+\sin \varphi_i\right) \cos \varphi}{(1+\sin \varphi) \cos \varphi_i}\right] \nonumber\]

Resolving in \(s\) the above, we find:

\[s(\varphi)=s_i+\frac{1}{2 D} \log \left[1+\frac{D\left(v_i \cos \varphi_i\right)^2}{g}\left(\frac{\sin \varphi_i}{\left(\cos \varphi_i\right)^2}-\frac{\sin \varphi}{(\cos \varphi)^2}+\log \left[\frac{\left(1+\sin \varphi_i\right) \cos \varphi}{(1+\sin \varphi) \cos \varphi_i}\right]\right)\right] \label{22}\]

Figure 7

From Figure 7 we can write:

\[{\overrightarrow{r}}_i=\overrightarrow{r}\left(s_i\right)\Rightarrow {\overrightarrow{r}}_{i+1}=\overrightarrow{r}\left(s_i+\mathrm{\Delta }s_i\right)=\overrightarrow{r}\left(s_i\right)+\frac{d\overrightarrow{r}}{ds}\left(s_i\right)\mathrm{\ }\mathrm{\Delta }s_i \label{23}\]

with \(\mathrm{\Delta }s_i=s_{i+1}-s_i\).

From the curvilinear coordinates, the Frenet-Serret formulae define the unit tangent vector \({\overrightarrow{u}}_T\) as:

\[\frac{d\overrightarrow{r}}{ds}={\overrightarrow{u}}_T=\left\langle {\mathrm{cos} \varphi \ },{\sin \varphi \ }\right\rangle \Rightarrow \frac{d\overrightarrow{r}}{ds}\left(s_i\right)={\left({\overrightarrow{u}}_T\right)}_i=\left\langle {\mathrm{cos} {\varphi }_i\ },{\sin {\varphi }_i\ }\right\rangle \nonumber\]

Therefore, Equation \ref{23} becomes:

\[{\overrightarrow{r}}_{i+1}={\overrightarrow{r}}_i+{\left({\overrightarrow{u}}_T\right)}_i\mathrm{\ }\mathrm{\Delta }s_i\Rightarrow \left(\ \genfrac{}{}{0pt}{}{x_{i+1}=x_i+{\mathrm{cos} {\varphi }_i\ }\left(\mathrm{\Delta }s_i=s_{i+1}-s_i\right)}{y_{i+1}=y_i+{\sin {\varphi }_i\ }\left(\mathrm{\Delta }s_i=s_{i+1}-s_i\right)}\ \right) \label{24}\]

The range \(x_R\) is determined numerically or graphically when:

\[\displaystyle \left\{ \begin{array}{c} \sum^{i=n}_{i=0}{y_{i+1}}=\sum^{i=n}_{i=0}{y_i+{\sin {\varphi }_i\ }\mathrm{\Delta }s_i}=0\to n \\ x_R=\sum^{i=n}_{i=0}{x_{i+1}}=\sum^{i=n}_{i=0}{x_{i+1}}x_i+{\mathrm{cos} {\varphi }_i\ }\mathrm{\Delta }s_i \end{array} \right. \nonumber\]

2.2.2: Graph

Another more analytic approach to generate the graph \(\overrightarrow{r}\left(\varphi \right)=\left\langle x\left(\varphi \right),y\left(\varphi \right)\right\rangle\), is to rewite Equation \ref{19} as follows:

\[\left\{\begin{array}{l}
\vec{\imath} \rightarrow-D v^{n-1}=\frac{d}{d t}\left(\log v_x\right) \\
\vec{\jmath} \rightarrow-\frac{g}{v_y}-D v^{n-1}=\frac{d}{d t}\left(\log v_y\right)
\end{array}\right. \label{25}\]

Subtracting Equation \ref{25}.2 to Equation \ref{25}.1, we get:

\[-\frac{g}{v_y}=\frac{d}{d t}\left(\log v_y\right)-\frac{d}{d t}\left(\log v_x\right)=\frac{d}{d t}\left[\log \left(\frac{v_y}{v_x}\right)\right]=\frac{d}{d \varphi}[\log (\tan \varphi)] \frac{d \varphi}{d t} \nonumber\]

\[-\frac{g \cos \varphi}{v}=\frac{d \varphi}{d t} \label{26}\]

Combining \ref{19}.1 and Equation \ref{26}, then we get:

\[-D v^{n-1} v_x=\frac{d v_x}{d t}=\frac{d v_x}{d \varphi} \frac{d \varphi}{d t}=-\frac{g \cos \varphi}{v} \frac{d v_x}{d \varphi} \nonumber\]

\[D\left(\frac{v_x}{\cos \varphi}\right)^{n-1} v_x=\frac{g(\cos \varphi)^2}{v_x} \frac{d v_x}{d \varphi} \Rightarrow \int_{v_{x_i}}^{v_x} \frac{d v_x}{v_x^{n+1}}=\frac{D}{g} \int_{\varphi_i}^{\varphi} \frac{d \varphi}{(\cos \varphi)^{n+1}} \nonumber\]

\[\frac{1}{n}\left(\frac{1}{v_{x_i}^n}-\frac{1}{v_x^n}\right)=\frac{D}{n g}\left(\frac{\sin \varphi}{(\cos \varphi)^n}+(n-1) \int \frac{d \varphi}{(\cos \varphi)^{n-1}}\right)_{\varphi_i}^{\varphi} \nonumber\]

Resolving the above in \(v_x\), we obtain:

\[v_x=\left(\left[\frac{1}{v_{x_i}^n}-\frac{D}{g}\left(\frac{\sin \varphi}{(\cos \varphi)^n}+(n-1) \int \frac{d \varphi}{(\cos \varphi)^{n-1}}\right)\right]_{\varphi_i}^{\varphi}\right)^{-\frac{1}{n}} \label{27}\]

In the other hand, from Equation \ref{26} we have:

\[d x=v_x d t=-\frac{v_x^2 d \varphi}{g(\cos \varphi)^2} \label{28}\]

Combining Equation \ref{27} and Equation \ref{28}, we obtain:

\[d x=-\frac{d \varphi}{g\left\{\left(\left[\frac{1}{v_{x_i}^n}-\frac{D}{g}\left(\frac{\sin \varphi}{(\cos \varphi)^n}+(n-1) \int \frac{d \varphi}{(\cos \varphi)^{n-1}}\right)\right]_{\varphi_i}^{\varphi}\right)^{\frac{1}{n}} \cos \varphi\right\}^2} \nonumber\]

And after integration we reach:

\[x(\varphi)=x_i-\frac{1}{g} \int_{\varphi_i}^{\varphi} \frac{d \varphi}{\left\{\left(\left[\frac{1}{v_{x_i}^n}-\frac{D}{g}\left(\frac{\sin \varphi}{(\cos \varphi)^n}+(n-1) \int \frac{d \varphi}{(\cos \varphi)^{n-1}}\right)\right]_{\varphi_i}^{\varphi}\right)^{\frac{1}{n}} \cos \varphi\right\}^2} \label{29}\]

In the \(\overrightarrow{j}\) direction we can write:

\[v_y=v_x \tan \varphi \Rightarrow y=y_i+\int v_x \tan \varphi d \varphi \label{30}\]

With Equation \ref{27} into Equation \ref{30}, we obtain:

\[y(\varphi)=y_i-\frac{1}{g} \int {\varphi_i}^{\varphi} \dfrac{\tan \varphi d \varphi}{\left\{\left(\left[\frac{1}{v_{x_i}^n}-\frac{D}{g}\left(\frac{\sin \varphi}{(\cos \varphi)^n}+(n-1) \int \frac{d \varphi}{(\cos \varphi)^{n-1}}\right)\right]_{\varphi_i}^{\varphi}\right)^{\frac{1}{n}} \cos \varphi\right\}^2} \label{31}\]

Therefore, from Equation \ref{29} and Equation \ref{31}, for \(n=2\) we have:

\[\left\{\begin{array}{l}
x_{i+1}=x_i+\int_{\varphi_i}^{\varphi_{i+1}} \frac{d \varphi}{\frac{g}{v_{x_i}^2}-D\left(\frac{\sin \varphi_i}{\cos ^2 \varphi_i}-\frac{\sin \varphi_{i+1}}{\cos ^2 \varphi_{i+1}}+\log \left[\frac{\left(1+\sin \varphi_i\right) \cos \varphi_{i+1}}{\left(1+\sin \varphi_{i+1}\right) \cos \varphi_i}\right]\right) \cos ^2 \varphi_{i+1}} \\
y_{i+1}=y_i+\int_{\varphi_i}^{\varphi_{i+1}} \frac{\tan \varphi d \varphi}{\frac{g}{v_{x_i}^2}-D\left(\frac{\sin \varphi_i}{\cos ^2 \varphi_i}-\frac{\sin \varphi_{i+1}}{\cos ^2 \varphi_{i+1}}+\log \left[\frac{\left(1+\sin \varphi_i\right) \cos \varphi_{i+1}}{\left(1+\sin \varphi_{i+1}\right) \cos \varphi_i}\right]\right) \cos ^2 \varphi_{i+1}}
\end{array}\right. \label{32}\]

The set of Equations Equation \ref{32} not only are more laborious than Equation \ref{24}, but less precises since with Equation \ref{24} the graph \(\overrightarrow{r}\left(\varphi \right)=\left\langle x\left(s\left(\varphi \right)\right),y\left(s\left(\varphi \right)\right)\right\rangle\) is generated form values analitically calculated by Equation \ref{22}, meanwhile the graph \(\overrightarrow{r}\left(\varphi \right)=\left\langle \ x\left(\varphi \right),\ y\left(\varphi \right)\ \right\rangle\) is generated form values numerically calculated by Equation \ref{32}.

Moreover, as we can see from Figure 8, the integrand functions of Equation \ref{32} have a singularity whenever:

\[\Phi_i(\varphi)=\frac{g}{D\left(v_i \cos \varphi_i\right)^2}+\frac{\sin \varphi_i}{\cos ^2 \varphi_i}+\log \left(\frac{1+\sin \varphi_i}{\cos \varphi_i}\right)-\left(\frac{\sin \varphi}{\cos ^2 \varphi}+\log \left(\frac{1+\sin \varphi}{\cos \varphi}\right)\right)=0 \nonumber\]

Therefore, the numerical integration of the integrand function of Equation \ref{32} in proximity of the singularity, becomes erratic.

Figure 8

Figure 9 shows the comparison of the results between Equation \ref{24} and Equation \ref{32}, Integrated with the Simpson’s 1/3 rule.

Figure 9

2.2.3 Graph

\(\overrightarrow{r}\left(t\right)=\left\langle \ x\left(t\right)\ ,\ y\left(t\right)\ \right\rangle\).

Let us expand the expression of the vector position \({\overrightarrow{r}}_{i+1}\left(t\right)\) with a Taylor series till the second order:

\[\vec{r}_{i+1}=\vec{r}_i\left(t_{i+1}\right)=\vec{r}_i\left(t_i+\Delta t\right)=\vec{r}_i+\frac{d \vec{r}_i}{d t} \Delta t+\frac{d^2 \vec{r}_i}{d t^2} \frac{\Delta t^2}{2}=\vec{r}_i+\vec{v}_i \Delta t+\frac{d \vec{v}_i}{d t} \frac{\Delta t^2}{2} \label{33.1}\]

In other terms

\[\left\langle x_{i+1}, y_{i+1}\right\rangle=\left\langle x_i, y_i\right\rangle+v_i\left\langle\cos \varphi_i, \sin \varphi_i\right\rangle \Delta t+\left\langle\frac{d \vec{v}_{x, i}}{d t}, \frac{d \vec{v}_{y, i}}{d t}\right\rangle \frac{\Delta t^2}{2} \label{33.2}\]

From the set Equation \ref{4} we get:

\[\left\{\begin{array}{l}
-D \frac{v_i^2}{d t} \cos \varphi_i=\frac{d v_{x, i}}{d t} \\
-g-\frac{v_i^2}{d t} \sin \varphi_i=\frac{d v_{y, i}}{d t}
\end{array}\right. \label{34}\]

Combining \ref{33.2} and Equation \ref{34}, we obtain:

\[\left\{\begin{array}{l}
x_{i+1}=x_i+v_i \cos \varphi_i \Delta t-\frac{D \cos \varphi_i\left(v_i \Delta t\right)^2}{2} \\
y_{i+1}=y_i+v_i \sin \varphi_i \Delta t-\left(D v_i^2 \sin \varphi_i+g\right) \frac{\Delta t^2}{2}
\end{array}\right. \label{35}\]

Equation \ref{26} under a discrete form gives the angles \({\varphi }_{i+1}\):

\[-\frac{g \cos \varphi}{v}=\frac{d \varphi}{d t} \rightarrow \frac{\Delta \varphi}{\Delta t}=\frac{\varphi_{i+1}-\varphi_i}{\Delta t}=-\frac{g \cos \varphi_i}{v_i} \nonumber\]

\[\varphi_{i+1}=\varphi_i-\frac{g \cos \varphi_i \Delta t}{v_i} \label{36}\]

Equation \ref{20} under a discrete form gives the velocity \(v_{i+1}\):

\[v_x=v_{x_i} e^{-D\left(s-s_i\right)} \rightarrow v_{i+1}=\frac{v_i \cos \varphi_i}{\cos \varphi_{i+1}} e^{-D\left(\Delta s_i=\sqrt{\left(\Delta x_i\right)^2+\left(\Delta y_i\right)^2}\right)} \nonumber\]

\[v_{i+1}=\frac{v_i \cos \varphi_i}{\cos \varphi_{i+1}} e^{-D\left(\sqrt{\left(x_{i+1}-x_i\right)^2+\left(y_{i+1}-y_i\right)^2}\right)} \label{37}\]

The time \(t_n\) is therefore: \[t_n=\sum^n_{i=1}{i}\mathrm{\Delta }t\mathrm{\ Equation \label{38}}\]

With Equations \ref{35}, \ref{36}, \ref{37}, and \ref{38}, starting from the initial conditions at \(i=0\), we calculate step by step all the data of interest.

The graph \(\overrightarrow{r}\left(t\right)=\left\langle \ x\left(t\right)\ ,\ y\left(t\right)\ \right\rangle\) is identical to \(\overrightarrow{r}\left(t\right)=\left\langle \ x\left(\varphi \right)\ ,\ y\left(\varphi \right)\ \right\rangle\) for the same initial conditions:

Figure 10.1

Figure 10.2

2.3: \(\boldsymbol{n} \in \mathbb{R}^{+}\)

Is the general case where experimental data suggest that \(n\) is not necessarily a natural number and can be any real positive number.

Equation \ref{33.2} states:

\[\left\langle x_{i+1}, y_{i+1}\right\rangle=\left\langle x_i, y_i\right\rangle+v_i\left\langle\cos \varphi_i, \sin \varphi_i\right\rangle \Delta t+\left\langle\frac{d \vec{v}_{x, i}}{d t}, \frac{d \vec{v}_{y, i}}{d t}\right\rangle \frac{\Delta t^2}{2} \nonumber\]

And the set of D.E. in Equation \ref{4} gives:

\[\left\{\begin{array}{l}
\vec{\imath} \rightarrow-D v^{n-1} v_x=\frac{d v_x}{d t} \\
\vec{\jmath} \rightarrow-g-D v^{n-1} v_y=\frac{d v_y}{d t}
\end{array}\right. \nonumber\]

Combining Equations \ref{33.2} and \ref{4}, we obtain:

\[\left\{\begin{array}{l}
x_{i+1}=x_i+v_{x, i} \Delta t-D v_i^{n-1} v_{x, i} \frac{\Delta t^2}{2} \\
y_{i+1}=y_i+v_{y, i} \Delta t-\left(D v_i^{n-1} v_{y, i}+g\right) \frac{\Delta t^2}{2}
\end{array}\right. \label{39}\]

We have to find an expression for \({\overrightarrow{v}}_{i+1}\). Let us expand the vector velocity \({\overrightarrow{v}}_{i+1}\left(t\right)\) with a Taylor series till the second order:

\[{\overrightarrow{v}}_{i+1}={\overrightarrow{v}}_i\left(t_i+\mathrm{\Delta }t\right)={\overrightarrow{v}}_i+\frac{d{\overrightarrow{v}}_i}{dt}\mathrm{\Delta }t+\left[\frac{d^2{\overrightarrow{v}}_i}{dt^2}=\frac{d}{dt}\left(\frac{d{\overrightarrow{v}}_i}{dt}\right)\right]\frac{{\mathrm{\Delta }t}^2}{2} \label{40}\]

In other terms:

\[\left\{\begin{array}{l}
v_{x, i+1}=v_{x, i}+\frac{d v_{x, i}}{d t} \Delta t+\frac{d}{d t}\left(\frac{d v_{x, i}}{d t}\right) \frac{\Delta t^2}{2} \\
v_{y, i+1}=v_{x, i}+\frac{d v_{y, i}}{d t} \Delta t+\frac{d}{d t}\left(\frac{d v_{y, i}}{d t}\right) \frac{\Delta t^2}{2}
\end{array}\right. \label{41}\]

Let us calculate \(\frac{d}{dt}\left(\frac{dv_{x,i}}{dt}\right)\). From \ref{4}.1 we can write:

\[\frac{d}{d t}\left(\frac{d v_{x, i}}{d t}\right)=-D \frac{d}{d t}\left(v_i^{n-1} v_{x, i}\right)=-D \frac{d}{d t}\left(\frac{v_{x, i}^n}{\cos ^{n-1} \varphi_i}\right) \nonumber\]

\[\frac{d^2 v_{x, i}}{d t^2}=-D\left(\frac{v_{x, i}^{n-1}}{\cos ^n \varphi_i}\right)\left((n-1) v_{x, i} \sin \varphi_i \frac{d \varphi_i}{d t}+n \cos \varphi_i \frac{d v_{x, i}}{d t}\right) \label{42}\]

With \ref{4}.1 and Equation \ref{26} into Equation \ref{42}, we get:

\[\frac{d^2 v_{x, i}}{d t^2}=-D\left(\frac{v_{x, i}^{n-1}}{\cos ^n \varphi_i}\right)\left(-(n-1) v_{x, i} \sin \varphi_i \frac{g \cos ^2 \varphi_i}{v_{x, i}}+n \cos \varphi_i\left[-D\left(\frac{v_{x, i}^{n-1}}{\cos ^n \varphi_i}\right)\right]\right) \nonumber\]

After the due simplifications, we obtain:

\[\frac{d^2 v_{x, i}}{d t^2}=-D v_i^{n-1}\left((n-1) g \sin \varphi_i \sin \varphi_i+n D v_i^{n-1}\right) \label{43}\]

Let us calculate now \(\frac{d}{dt}\left(\frac{dv_{y,i}}{dt}\right)\). From Equation \ref{4}.2 we can write:

\[\frac{d}{d t}\left(\frac{d v_{x, i}}{d t}\right)=-\frac{d}{d t}\left(g+D v_i^{n-1} v_{y, i}\right)=-D \frac{d}{d t}\left(\frac{v_{y, i}^n}{\sin ^{n-1} \varphi_i}\right) \nonumber\]

\[\frac{d^2 v_{y, i}}{d t^2}=-D v_i^{n-1}\left((1-n) \frac{v_{y, i}}{\tan \varphi_i} \frac{d \varphi_i}{d t}+n \frac{d v_{x, i}}{d t}\right) \label{44}\]

With Equation \ref{4}.2 and Equation \ref{26} into Equation \ref{44}, we get:

\[\frac{d^2 v_{y, i}}{d t^2}=D v_i^{n-1}\left[-(1-n) \frac{v_{y, i}}{\tan \varphi_i} \frac{g \cos ^2 \varphi_i}{v_{x, i}}-n\left(g+D v_i^{n-1} v_{y, i}\right)\right] \nonumber\]

After the due simplifications, we obtain:

\[\frac{d^2 v_{y, i}}{d t^2}=D v_i^{n-1}\left[(1-n) g \cos ^2 \varphi_i+n\left(g+D v_i^{n-1} v_{y, i}\right)\right] \label{45}\]

With Equation \ref{43} and Equation \ref{45} into Equation \ref{41}, we get the set of Equations for the velocities:

\[\left\{\begin{array}{l}
v_{x, i+1}=v_{x, i}+D v_i^{n-2}\left\{v_i\left[(n-1) g \sin \left(2 \varphi_i\right) \Delta t-4 v_{x, i}\right]+2 n D v_i^n v_{x, i}\right\} \frac{\Delta t}{4} \\
v_{y, i+1}=v_{y, i}-g \Delta t+D v_i^{n-2}\left\{v_i\left(g\left[n+(1-n) \cos ^2 \varphi_i\right] \Delta t-2 v_{y, i}\right)+n D v_i^n v_{y, i} \Delta t\right\} \frac{\Delta t}{2} \\
v_i=\sqrt{v_{x, i}^2+v_{y, i}^2}
\end{array}\right. \]

Equation \ref{26} under a discrete form gives the angles \({\varphi }_{i+1}\):

\[-\frac{g \cos \varphi}{v}=\frac{d \varphi}{d t} \rightarrow \frac{\Delta \varphi}{\Delta t}=\frac{\varphi_{i+1}-\varphi_i}{\Delta t}=-\frac{g \cos \varphi_i}{v_i} \nonumber\]

\[\varphi_{i+1}=\varphi_i-\frac{g \cos \varphi_i \Delta t}{v_i} \nonumber\]

The time \(t_n\) is therefore:

\[t_n=\sum_{i=1}^n i \Delta t \nonumber\]

With Equations \ref{33.2}, \ref{36}, \ref{38}, and \ref{46}, starting from the initial conditions at \(i=0\), we calculate step by step all the data of interest.

Figure 11 shows the comparative graph of the family of trajectories for different values of \(n\).

Figure 11

Figure 12 shows the graph of the distribution of the velocities for \(n=1.6\).

Figure 12

Note

In the graphs of the distribution of the velocities, the term \(v_t\) is the “terminal velocity” calculated from 4.1) when \(\varphi \to -\frac{\pi }{2}\), i.e. the free fall:

\[\left\{\begin{array}{l}
\vec{\imath} \rightarrow-D v^{n-1}\left(v_x=0\right)=0 \\
\vec{\jmath} \rightarrow-g-D v_t^{n-1}\left(v_y=\frac{v_t}{\sin \left(-\frac{\pi}{2}\right)}\right)=-g+D v_t^n=0
\end{array} \Rightarrow v_t=\left(\frac{g}{D}\right)^{\frac{1}{n}}\right. \nonumber\]

 Figure 13 shows the comparative graph of the trajectories calculated with above different Equations.

Figure 13