It looks like the Mythbusters tend to ignore air resistance.

In a recent episode, they claimed to have demonstrated that a horizontally fired bullet and a bullet that is simply dropped fall to the ground in the same amount of time. They were wrong. What they actually demonstrated is that air resistance causes the fired bullet to hit the ground more slowly.

Their argument would apply perfectly in a vacuum, as on the surface of the Moon, but not here on the Earth, where the bullet’s motion is governed not just by the laws of gravity, but also by the laws of a non-conservative force, namely air resistance. (Why is it non-conservative? Some of the bullet’s kinetic energy is converted into heat, as it travels through the air at high speed. Unless we also include the thermodynamics of the air into our equations of motion, the equations will not conserve energy, as the amount of kinetic energy converted into heat will just appear “lost”.)

The bullet’s velocity, $v$, can be written as $v^2=v_h^2+v_v^2$, where $v_h$ is the horizontal and $v_v$ is the vertical component. The initial horizontal velocity is $v_0$. The initial vertical velocity is 0.

Air resistance is proportional to the square of the bullet’s velocity. To be precise, acceleration due to air resistance will be

$a_\mathrm{air}=\kappa v^2,$

where $\kappa$ is an unknown proportionality factor. (To be more precise, $\kappa=\frac{1}{2}cA\rho m$, where $c$ is the dimensionless drag coefficient, $A$ is the bullet’s cross-sectional area, $\rho$ is the density of the air, and $m$ is the bullet’s mass.) The direction of the acceleration will be opposite the direction of the bullet’s motion.

The total acceleration of the bullet will have two components: a vertical component $g$ due to gravity, and a component $a_\mathrm{air}$ opposite the direction of the bullet’s motion.

The resulting equations of motion can be written as:

\begin{align}\frac{dv_h}{dt}&=-\kappa vv_h,\\\frac{dv_v}{dt}&=g-\kappa vv_v.\end{align}

Right here we can see the culprit: air resistance not only slows the bullet down horizontally, it also reduces its downward acceleration.

This is a simple system of two differential equations in the two unknown functions $v_h(t)$ and $v_v(t)$. Its solution is not that simple, unfortunately. However, it can be greatly simplified if we notice that given that $v_v\ll v_h$, $v\simeq v_h$, and therefore, we get

\begin{align}\frac{dv_h}{dt}&=-\kappa v_h^2,\\\frac{dv_v}{dt}&=g-\kappa v_vv_h.\end{align}

This system is solved by

\begin{align}v_h&=\frac{1}{\kappa t+C_1},\\v_v&=v_h\left[\left(\frac{1}{2}\kappa t^2+C_1t\right)g+C_2\right].\end{align}

Given

$v_0=\frac{1}{\kappa t_0+C_1},$

we have

$C_1=\frac{1}{v_0-\kappa t_0},$

$v_h=\frac{v_0}{\kappa(t-t_0)v_0+1},$

or, if we set $t_0=0$,

$v_h=\frac{v_0}{\kappa tv_0+1}.$

Similarly, given $v_v(0)=0$, we get $C_2=0$, thus

$v_v=gt\frac{\kappa v_0t+2}{2\kappa v_0t+2}.$

The distance traveled horizontally ($s_h$) and vertically ($s_v$) between $t_0=0$ and $t_1$ can be obtained by simple integration of the respective velocities with respect to $t$ between 0 and $t_1$:

\begin{align}s_h&=\frac{\log(\kappa v_0t_1+1)}{\kappa},\\s_v&=g\frac{\kappa v_0t_1(\kappa v_0t_1+2)-2\log(\kappa v_0t_1+1)}{(2\kappa v_0)^2}.\end{align}

The claim by the Mythbusters was that the time it took for the fired bullet to hit the ground was only $\sim 40~{\rm ms}$ more than the time it took for a dropped bullet to fall, which is a negligible difference. But it is not! Taking $g\simeq 10~{\rm m}/{\rm s}^2$, it is easy to see that the time it takes for a bullet to fall from a height of $1~{\rm m}$, using the well-known formula $\frac{1}{2}gt^2$, is $447~{\rm ms}$; the difference measured by the Mythbusters is nearly 10% of this number!

Not only did the fired bullet take longer to hit the ground, the Mythbusters’ exquisite setup allows us to calculate the bullet’s initial velocity $v_0$ and drag coefficient $\kappa$. This is possible because the Mythbusters conveniently provided three pieces of information (I am using approximate numbers here): the length of the path that the bullet traveled $s_h\simeq 100~{\rm m}$, the height of the bullet at the time of firing ($s_v\simeq 1~{\rm m}$), and the time it took for the fired bullet to hit the ground. Actually, what they provided was the difference between the time for a fired vs. a dropped bullet to hit the ground, but we know what it is for the dropped bullet (and because it is never moving very rapidly, we can ignore air resistance in its case), so $t_1=447+40=487 {\rm ms}$. The solution is given by

\begin{align}\kappa&=0.0054~\mathrm{m}^{-1},\\v_0&=272.3~\mathrm{m}/\mathrm{s}.\end{align}

Given a bullet cross-sectional area of $A=2~{\rm cm}^2=2\times 10^{-4}~{\rm m}^2$, an approximate air density of $\rho=1~{\rm kg}/{\rm m}^3$, and a bullet mass of $m=20~{\rm g}=0.02~{\rm kg}$, the dimensionless drag coefficient for the bullet can be calculated as $c=2\kappa mA\rho=1.08$, which is not at all unreasonable for a tumbling bullet. Of course the actual values of $A$ and $m$ may differ from the ones I’m using here, resulting in a different value for the dimensionless drag coefficient $c$.

And now (May 28, 2012) I feel obliged to include a footnote, given the almost hysterical news coverage in the last couple of days about an exact solution to this same problem obtained by a 16-year old Dresden youth of Indian descent, Shouryya Ray.

The news coverage was breathless but incomplete: the youth, we were told, solved a problem that baffled Newton 350 years ago, but there was no description of the actual problem (other than a hint that it had to do with air resistance) and its solution, except for a press photo with him holding up a large sheet of paper with an equation.

I suspected that his variables $u$ and $v$ were in fact the horizontal and vertical components of a projectile, which I denoted with $v_h$ and $v_v$ above. So Ray's equation would read, in my notation:

$\frac{g^2}{2v_h^2}-\frac{\kappa g}{2}\left(\frac{v_v\sqrt{v_h^2+v_v^2}}{v_h^2}+{\rm arcsinh}\left|\frac{v_v}{v_h}\right|\right)={\rm const,}$

where I picked up an extra minus sign because I considered $g$ to be positive when pointing downwards.

But how did he find this neat form? Is it even correct? To find out, I tried a few ways to integrate the equations but to no avail. I admit I was ready to give up when I came across a solution posted on reddit. The derivation is actually quite simple. We begin with (I now use the overdot for the time derivative):

\begin{align}
\dot{v}_h=-\kappa\sqrt{v_h^2+v_v^2}v_h,\\
\dot{v}_v=g-\kappa\sqrt{v_h^2+v_v^2}v_v.
\end{align}

Now multiply the first equation by $\dot{v}_v$ and the second by $\dot{v}_h$:

\begin{align}
\dot{v}_h\dot{v}_v=-\kappa\sqrt{v_h^2+v_v^2}v_h\dot{v}_v,\\
\dot{v}_h\dot{v}_v=g\dot{v}_h-\kappa\sqrt{v_h^2+v_v^2}\dot{v}_hv_v.
\end{align}

Subtract one from the other and rearrange:

$g\dot{v}_h=\kappa\sqrt{v_h^2+v_v^2}(\dot{v}_hv_v-v_h\dot{v}_v).$

Now substitute $v_v=sv_h$:

$g\dot{v}_h=\kappa\sqrt{(s^2+1)v_h^2}(\dot{v}_hv_v-v_h\dot{v}_v).$

Divide both sides by $v_h^3$:

$g\frac{\dot{v}_h}{v_h^3}=-\kappa\sqrt{(s^2+1)}\dot{s}.$

In this form, both sides are full derivatives with respect to $t$ and thus both sides can be integrated, to yield

$-\frac{1}{2}\frac{g}{v_h^2}=-\kappa\left[\frac{1}{2}s\sqrt{s^2+1}+\frac{1}{2}{\rm arcsinh}~s\right]+C,$

where $C$ is an integration constant. Multiplying by $g$ and replacing $s$ with $v_v/v_h$, we get back Ray's equation.

This is an elegant result. Its practical utility may be limited, however, as the solution is implicit. Actually computing $v_h$ and $v_v$ as functions of time still requires numerical methods, so one might as well just solve the original differential equations numerically. Perhaps this explains why Ray's formula is not better known. But it has been discovered before. Apart from trivial notational differences, this very formulation is discussed by Parker (Am. J. Phys, 45, 7, 606, July 1977). But given its importance to ballistics, it should come as no surprise that this problem is discussed in depth in textbooks dating back to the middle of the 19th century, for instance in an 1860 book by Didion.