Q:

What is the terminal velocity for a mobile phone

You may have seen the story of the iPhone which was dropped from perhaps 13,500 feet by a skydiver - it survived.
This made me wonder how to work out the terminal velocity for something like that. Obviously calculating terminal velocity for a sphere can be relatively straightforward, but with a flattened oblong, what factors come into play?
End on will be fast, flat will be slow, but is there a stable configuration?

A:

From the equilibrium between drag and weight:
$$
\frac{1}{2} C_x \rho v^2 S = m g
$$
we can write the terminal velocity as
$$
v = \sqrt{\frac{2 m g}{C_x \rho S}}
$$
where $m$ is the mass of the phone, $C_x$ its drag coefficient,
$S$ its section, $g$ the acceleration of gravity,
and $\rho$ the density of air.
Now, this is not really a good answer, as the big question is how to
estimate $S$ (depends on the phone orientation) and $C_x$. But at least
you can compute an order of magnitude, as in most cases $C_x$ of of
order 1.

