# Homework Help: Air resistance & Trajectory

1. Aug 17, 2005

### bomba923

Mathematically, how would air resistance affect trajectory?
Suppose I shoot a cannonball with initial velocity $v_0$ at an angle of elevation $$\theta$$. Air resistance is a force antiparallel to velocity, represented by the equation (according to Barron's Physics C review book!) $\vec F_{air} = - c \cdot \vec v$, where $c$ is a constant SI-expressed in kg/sec.
Because the direction of velocity changes with $t$, so does, as well, the direction of air resistance. To represent its effect in the x & y directions, let $\alpha$ represent the instantaneous angle that $\vec v$ makes with the x-axis. Thus, using 2D Cartesian, I represent the velocities as
$$\left\{ \begin{gathered} \vec v_x = \left| {v_0 } \right|\cos \theta - t\left( {c \cdot \vec v} \right)\cos \alpha \hfill \\ \vec v_y = \left| {v_0 } \right|\sin \theta - t\left( {\vec g + c \cdot \vec v \cdot \sin \alpha } \right) \hfill \\ \end{gathered} \right\}$$
*But how do I find y(t) and x(t) ? ?

Last edited: Aug 17, 2005
2. Aug 17, 2005

### Tide

I think the equations you are looking for are:

$$\frac {dv_x}{dt} = -c v_x$$

$$\frac {dv_y}{dt} = -g - c v_y$$

Try solving them for $v_x$ and $v_y$ and then integrating to find x and y.

3. Aug 17, 2005

### Tide

Try letting $w = v_y + \frac {mg}{c}$. That should simplify things. Don't forget to transform back to $v_y$ when you're done.

4. Aug 18, 2005

### bomba923

(Hehe-this got moved to homework "College level" yet this is not homework and I'm just a curious HS student)
Anyway, both differentials are simple enough, so considering mass properly (b/c initially I considered it improperly):
*Since $v_{x,0} = v_0 \cos \theta$,
$$m\frac{{dv_x }}{{dt}} = - cv_x \Rightarrow v_x = v_0 e^{ - ct/m} \cos \theta$$
*Since $v_{y,0} = v_0 \sin \theta$,
$$m\frac{{dv_y }}{{dt}} = - mg - cv_y \Rightarrow v_y = \frac{m}{c}\left[ {\left( {g + cv_0 m^{ - 1} \sin \theta } \right)e^{ - ct/m} - g} \right]$$
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Hmm...will trajectories affected by air resistance in the form $\vec F_{air} = c \cdot \vec v$ deviate from a parabolic shape? I integrated both $v_x$ and $v_y$ (where $c \ne 0$) with respect to time...and it appears that the trajectory deviates from a parabolic shape...hmm..

Last edited: Aug 18, 2005