For a ball of mass $m$, radius $r$, falling freely through air of density $p$, the force of air resistance is proportional to the density of the air, the square of the ball's speed, $v$, and the area of a projection of the ball onto a plane perpendicular to the direction of its motion. Thus, in one dimension, the equation of motion is:

$mx'' = mg - 0.5CApv^2$ , where g is the acceleration due to gravity

The factor of $0.5$ is just for convention. If the motion is considered in two dimensions instead, then the following equations describe it:

$mx'' = -0.5CApv_xv$

$my'' = mg - 0.5CApv_yv$

With v = $(v_x,v_y)$ and $v = √(v_x^2 + v_y^2)$

Now my question is: why do the terms $v_xv$ and $v_yv$ appear in the above equations, instead of $v_x^2$ and $v_y^2$ respectively, in accordance with the 1-dimensional case?

My attempt was to realize that the area of the projection , $A$, is different when the motion is split into two dimensions. Then I tried to consider the horizontal and vertical components of the area, to no avail.I also thought that, for instance in the $x$ direction, $v_y = 0$ so $v =√(v_x^2) = v_x$. So the term $v_xv$ reduces to $v_x^2$. Is there some other explanation?

  • 1
    $\begingroup$ As an aside (not really important for this particular question), the factor of $0.5$ is more than just a convention. It comes from the formula for dynamic pressure, which is $\frac12 \rho v^2$, and the $\frac12$ occurs in that formula for the same reason it occurs in the formula for kinetic energy of a particle of mass $m$, $\frac12 mv^2$. $\endgroup$
    – David K
    Sep 27, 2015 at 15:26
  • $\begingroup$ Aha, thanks once again. $\endgroup$
    – 2good4this
    Sep 27, 2015 at 15:43

1 Answer 1


The magnitude of the force of air resistance on the ball is still $D = \frac12 C_D A\rho v^2$, the same as for one-dimensional motion. In fact it must be, because you could construct a coordinate axis parallel to the direction of motion at that instant, and you would (at least at that instant) have one-dimensional motion in that direction.

What changes is that since we have put coordinate axes that are not parallel to the direction of motion, we want to find the components of the $D$ force parallel to each coordinate axis. The force triangle is similar to the velocity triangle (same angles with each axis). So since the velocity $v$ breaks up into components this way, \begin{align} v_x &= \left(\frac{v_x}{v}\right) v, \\ v_y &= \left(\frac{v_y}{v}\right) v, \end{align} the force $D$ breaks up into components in the same proportions, but in the opposite direction: \begin{align} D_x &= -\left(\frac{v_x}{v}\right) D, \\ D_y &= -\left(\frac{v_y}{v}\right) D. \end{align}

Plug in the known value of $D$ and work it out: \begin{align} D_x &= -\left(\frac{v_x}{v}\right) \frac12 C_D A\rho v^2 = -\frac12 C_D A\rho v_x v, \\ D_y &= -\left(\frac{v_y}{v}\right) \frac12 C_D A\rho v^2 = -\frac12 C_D A\rho v_x v. \end{align}

  • $\begingroup$ Briliiant! Thank you David. $\endgroup$
    – 2good4this
    Sep 27, 2015 at 15:30

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