You know who's better at guitar than he lets on? John freaking Mayer. Let's get started.
Getting to what I hope will become the "meat" of this blog - science and technology - I'd like to take some time to discuss gravity. Gravity is counted among the four fundamental interactions (alongside electromagnetism, the weak interaction and the strong interaction). It is similar to the electromagnetic interaction in that its magnitude is inversely proportional to distance squared, but different in that it is never repulsive. It is also the weakest of the four interactions, and is only noticeable when very massive objects are present (think planets). I won't go into the relativistic explanation of gravity (yet) - just good old fashioned Newtonian gravity.
$$F_G = G{M\cdot m\over{r^2}}$$
"G" (the Gravitational constant) and "M" (the mass of Azeroth) are constants, and "r" varies very little so long as our maximum height is only a tiny fraction of Azeroth's radius, so we can work with the highschool-physics friendly
$$F_G = mg$$
Normal mode: Constant acceleration with kinematics
Ultimately, I'd like to solve for "g" to compare gravity on Azeroth to gravity on Earth. This is easiest if the effect of air resistance is negligible. At first, I'll treat this as a constant acceleration problem. We can start with the parametric set of equations
$$x(t) = v_it$$
$$y(t) = h_0 - {1\over2}at^2$$
Note: these equations (called kinematics equations) are derived using constant acceleration. For systems with non-constant acceleration, integration or energy relations should be used.
Because the time and height are measurable, the first equation in that system isn't strictly necessary (but I'll keep it around for when I take measurements). Because "h0" (the initial height) is defined relative to the ground, we can say y(t) (the current height) is zero when an object hits the ground. Rearranging the second equation with this in mind gives
$$a = {2h_0\over{t^2}}$$
The solution contains two independent variables which can be measured in a single trial. Alternatively, if horizontal distance proves to be a more reliable measurement than time, we can use
$$a = {2h_0v_i^2\over{x^2}}$$
Easy enough, right?
Hard mode: Non-constant acceleration with differential equations
Still assuming that gravity is near-constant near Azeroth's surface, we'll look at a falling body experiencing drag. Drag (air resistance) is a force which is proportional to the magnitude of velocity squared, but opposite in direction. The full drag equation is:
$$F_D = {1\over2}{\rho}v^2C_DA$$
I'd like to take the time to define terminal velocity, because it will make our solutions much neater. With regards to "free fall", terminal velocity is the velocity at which FD = FG - that is to say that the force of drag (which is proportional to velocity squared) is equal to the force of gravity (which is constant). When these two forces are equal, no acceleration occurs; this is why skydivers don't accelerate indefinitely. Terminal velocity is related to mass and gravitational acceleration by the following equation:
$$v_\infty = \sqrt{\frac{2mg}{\rho C_DA}}$$
and makes acceleration due to drag easier to define:
$$a_D = \frac{gv^2}{v_\infty^2}$$
To my knowledge, there is no neat equation which describes an object experiencing a constant force and drag when the initial velocity is not parallel with the force. For the sake of readability (and my sanity), we're going to take the x-direction out of consideration here and simply fall. Through the use of
$$a = g - \frac{gv^2}{v_\infty^2}$$
becomes
$$0 = h_0 - \frac{v_\infty^2}{g}\ln{\cosh{\frac{gt}{v_\infty}}}$$
We have four variables to deal with: two (height and time) can be measured, but the others (gravitational acceleration and terminal velocity) must be solved for - meaning we'll have to take measurements from at least two different heights.
Stay tuned for "Gravity on Azeroth, part 2", in which I actually take measurements (read: jump off tall things for science)!
