What is the terminal velocity of a sheep? Inspired by this question on Gaming.SE
Using actual in-real-life physics, what would the terminal velocity of a sheep actually be? I would assume it would be around 50m/s, but I might be wrong.
Bonus question:
What would the terminal velocity of a chicken be?
Animal-friendly answers preferred.
 A: Like humans, the sheep has some – although more limited – freedom to act if it wants to change the asymptotic speed.
A skydiver's asymptotic speed may be between 50 m/s in the "face down" free fall and 90 m/s when he or she pulls the limbs in. These two speeds hugely differ which indicates that it doesn't make much sense to talk about a "universal" value, at least not too accurately.
The speed around 90 m/s is close to the terminal speed of any macroscopic animal that works hard to increase the terminal speed as much as it can. I've mentioned that a skydiver may approach this speed if he tries. But it is also the maximum speed of the falcons when they are hunting their prey.
Chickens are not predators but they won't be too far from that. Sheep will be a little bit slower but not much – the sheep are probably closer to the skydiver with his limbs pulled in.
The Minecraft produces the 80 m/s terminal speed for related animals and my guess is that all such physics facts are tuned to rather realistic values in that game.
A: That's a very hard question to answer theoretically because the aerodynamic drag, and therefore the terminal velocity, is highly dependant on the shape of the falling object.
Assuming we're in the turbulent regime, the aerodynamic drag is given by the equation:
$$ F_d = \tfrac{1}{2}\rho v^2 C_d A $$
where $\rho$ is the density of the air, $v$ is the velocity, $A$ is the cross sectional area in the direction of motion and $C_d$ is the drag coefficient. The terminal velocity is just the velocity at which the drag force is equal to the weight $mg$, and a quick rearrangement gives:
$$ v_t = \sqrt{\frac{2mg}{\rho C_d A}} $$
In the UK a popular breed is the Texel, which weighs around 80kg. I can't find any info on their size, but from my memories of sheep I'd guess they're about a metre long and the diameter of the body is about half a metre. The legs are pretty spindly, so the cross sectional area is going to be around $A = \pi/16$ m$^2$ if the sheep is falling head first or around 0.5 m$^2$ if it's falling sideways. Putting in the relevant figures we get:
$$ v_t \approx \frac{80}{\sqrt{C_d}} $$
The problem is what value to use for $C_d$. A sphere has $C_d = 0.47$, which would give $v \approx 117$ m/sec. But a cylinder (with sharp rims) has $C_d = 0.82$, which would give $v \approx 88$ m/sec. I'd guess the truth is somewhere in between.
Both these figures are higher than the terminal velocity of a human falling, which varies from around 60 to 90 m/sec depending on your orientation. But then sheep are pretty dense. If you've ever tried to pick one up you were probably surprised by how heavy they seem for their size. Bear in mind the Texel sheep I used as an example weighs 10kg more than I do and I'm 5'10". It's quite reasonable that the terminal velocity of a sheep would be higher than that of a human.
