Terminal Velocity of 100lb Hail In the Apocalypse there is meant to be hail weighing 100 pounds or 45 kg.
What would the terminal velocity be, assuming the hailstones are perfect spheres of solid ice?
Also, is there a way to guess the terminal velocity if the hailstones were not perfect spheres (and more like real hailstones)?
 A: Air drag is given by
$$F_d = \frac12 \rho v^2 A C_D$$
where $\rho$ is the density of air (variable, depending on temperature etc - 1.2 kg/m$^3$ is a reasonable approximation), $v$ is the velocity, $A$ is the projected area, and $C_D$ is the drag factor - which is a function of shape and Reynolds number. 0.5 is an OK approximation but it depends on the velocity - see this graph (from http://www.islandone.org/LEOBiblio/SPBI1GU5.GIF).

The terminal velocity is reached when the force is equal to the weight, $m g$.
We get $A$ from the mass and density. Density of ice is about 920 kg/m^3, so a 45 kg sphere of ice has a radius $r$ found by solving
$$\frac43 \pi r^3 \rho_{ice} = m_{ice}\\
r = \sqrt[3]{\frac{3 m_{ice}}{4\pi}}=0.23 m$$
This makes the projected area $A=\pi r^2 =  0.16 m^2$. So the terminal velocity is found by solving for $v$, and our initial estimate is:
$$v = \sqrt{\frac{2 m g}{\rho A C_D}}=\sqrt{\frac{2*45*9.8}{1.2*0.16*0.5}}=100 \;\rm{m/s}$$
If the hailstone is irregular shape, its terminal velocity will change a bit - but not a lot. Depending on the shape, it can be lower or higher.
Now that we have the velocity, we can look at the Reynolds number. A sphere of 0.23 m radius moving at 100 m/s in air has a Reynolds number of about $1.6\cdot 10^6$, and according to the above graph that results in a much lower $C_D$ of approximately 0.2. Repeating the calculation with that lower drag coefficient we find v = 150 m/s. In principle, we could compute a new R, and repeat... but I don't think the accuracy of the estimate warrants that.
150 m/s is 50% faster than the initial estimate! Note that I rounded my results - precisely because these kinds of errors quickly dominate the result, and giving an answer like "151.6 m/s", (which is what my calculator came up with), gives a false impression that the answer is not 151.7 or 151.5 - when in fact, a lot of estimating went into this and 150 is about as good as you can get.
A: According to WolframAlpha, $100\text{lbs}$ of ice has a volume $V=0.04536 \text{m}^3$ and mass $m=45.36\text{kg}$. We can find the radius of the ice ball:
$$
V=\frac{4}{3}\pi r^3 \rightarrow r = \sqrt[3]{\frac{3V}{4\pi}} = 0.221\text{m} = 22.1\text{cm}
$$
The force of air resistance is given by the equation:
$$
f_\text{drag}=-\frac{1}{2}C\rho Av^2
$$
Where $\rho$ is the density of air, $\approx 1.225 \text{kg}/\text{m}^3$, assuming room temperature air at sea level, $A$ is the cross sectional area of the sphere $=\pi r^2$, $v$ is the speed, and $C$ is the drag coefficient, $=0.5$ for a spherical object.
For a falling object, terminal velocity is reached when the upward force of air resistance exactly balances the downward force of gravity. There is no net force on the object and it stop accelerating:
$$
F_\text{net}=mg-\frac{1}{2}C\rho \pi r^2v^2 =0
$$
Solving for $v_\text{terminal}$:
$$
v_\text{terminal} = \sqrt{\frac{2mg}{C\rho\pi r^2}}
$$
Plug in values and solve:
$$
v_\text{terminal} = \sqrt{\frac{2(45.36\text{kg})(9.81\text{m}/{s}^2)}{(0.5)(1.225 \text{kg}/\text{m}^3) (0.221\text{m})^2\pi}} \approx 97.3\text{m}/\text{s}
$$
That's about $217\text{mph}$.
Solving for an irregular chunk of ice is exactly the same. The only things that change are the cross sectional area and the drag coefficient. Drag coefficients are usually found experimentally except for very simple shapes. I suspect that for a mostly spherical hail, the drag coefficient will be roughly the same. Another important thing to consider is the orientation of the hail. If the shape of the hail causes it to tumble as it falls, the drag coefficient and cross sectional area change is it rotates. Again I would suspect that for a mostly spherical hail it would not do any of these things. The short answer is that finding the drag characteristics of irregular objects is a painful process analytically. It is usually done experimentally or through computer modeling. The closest approximation is to assume that the hail is spherical.
