Air friction

A stone of mass 100 g is thrown with an intitial velocity $|\vec{v}|$ at an angle $\theta$ from horizontal. The stone is released from position $\vec{r}=0$. If air friction is ignored, the equation that describes the motion of the ball is,

$\vec{r} = vt\cos\theta \,\hat{x}+ \left( vt\sin\theta -\frac{gt^2}{2}\right)\,\hat{y}\,\text{m},$

where $g = 9.81$ m/s² is the acceleration of gravity at the earth's surface and $t$ is measured in seconds. The stone follows a parabola and returns to the height from which it was thrown, $y=0$, at a distance $x$ from where it was thrown. For a given initial velocity, the angle that maximizes the distance $x$ is $\theta = \frac{\pi}{4}\,\text{rad} = $ 45°.

If a linear friction force is included in the problem $\vec{F}_{drag} = -b\vec{v}$, then the optimal angle changes. The form below calculates the trajectory of the stone for different angles, friction contants, and intitial velocities.

$\vec{F} = m\frac{d^2\vec{r}}{dt^2} = -b\vec{v}-mg \,\hat{z}$

$b=$  N s/m

The initial conditions at time $t=0$ are $\vec{r}=0$
$|\vec{v}| =$  m/s  $\theta=$  deg

 3-D motion differential equation solver 

$ F_x=$

 [N]

$ F_y=$

 [N]

$ F_z=$

 [N]

$ m=$

 [kg]  
Initial conditions: 

$t_0=$

 [s] 

$\Delta t=$

 [s]

$x(t_0)=$

 [m] 

$N_{steps}$

$v_x(t_0)=$

 [m/s] 

Plot:

vs.

$y(t_0)=$

 [m] 

$v_y(t_0)=$

 [m/s] 

$z(t_0)=$

 [m] 

$v_z(t_0)=$

 [m/s] 

 
 

= , =


the animation to zoom or rotate.

 $t$ [s] $x$ [m] $y$ [m] $z$ [m] $v_x$ [m/s] $v_y$ [m/s] $v_z$ [m/s] $F_x$ [N] $F_y$ [N] $F_z$ [N] $P$ [W] $E_{\text{kin}}$ [J] $W$ [J]