Okay, something useful. Maybe.
Have you ever wondered how fast that rock at the top of the hill would be moving when it reached the bottom of the hill, if you were to push it off? Or how about how fast an object is moving just before it hits the ground? There's a pretty easy way to answer both these questions: conservation of energy. You've probably at least heard of it before, if you haven't encountered it in a physics course. There are two types of mechanical energy that would need to be considered in the above scenarios: potential energy (PE) and kinetic energy (KE). For the purposes of this discussion, I'll ignore non-mechanical energy. PE is the energy an object has due to its height above an arbitrary surface. For example, if there was a book on a shelf 5 feet high, then the potential energy would be equal to the energy the book would have if it fell off the shelf. This is a result of the conservation of energy, which simply means that energy in a closed system can not be created or destroyed. In this case, the closed system is the book. Since we are only considering mechanical energy, the book will only have PE and KE. Thus, the sum of the PE and KE of the book at any point in time will be equal to the sum of the PE and KE of the book at any other point in time. Expressed mathematically, PE_1 + KI_1 = PE_2 + KE_2.
Both PE and KE have relatively simple equation. PE is dependent on the mass of the object, its height above the ground or any other arbitrary surface, and the acceleration due to gravity. Expressed mathematically, PE = mgh. KE is a function of the object's mass and its velocity squared, or KE = 0.5mv^2.
Of course, the theory doesn't make a whole lot of sense until it's demonstrated, so I'll solve the problem with the book. We can state that mgh_1 + 0.5m(v_1)^2 = mgh_2 + 0.5m(v_2)^2. The mass of the book doesn't change, so all of the m's are equal and cancel out. This is extremely useful, because we don't even need to know the book's mass. The book is 5 feet off the ground, so h_1 = 5 ft and h_2 = 0 ft. If you wanted to, you could say something like h_1 = 15.72 ft and h_2 = 10.72 ft. It won't make any difference in the end. However, it's a lot easier to define the ground as being 0 ft. If we do that, we can immediately see that the term mgh_2 will just be 0, since any number multiplied by 0 is 0. We can also eliminate the term 0.5m(v_1)^2 because the book is just sitting on the shelf, so its initial velocity is 0. The acceleration due to gravity on earth is 32 ft/s^2 (9.8 m/s^2). This is all the information we need to solve the problem.
mgh_1 + 0.5m(v_1)^2 = mgh_2 + 0.5m(v_2)^2
Mass cancels.
gh_1 + 0.5(v_1)^2 = gh_2 + 0.5(v_2)^2
Substitute v_1 = 0 and h_2 = 0.
gh_1 = 0.5(v_2)^2
Solving for v_2.
v_2 = sqrt(2gh_1)
Substituting g = 32 ft/s^2 and h_1 = 5 ft.
v_2 = sqrt(2*(32 ft/s^2)*(5 ft)) = 17.9 ft/s.
This is the final solution. If the book falls off the shelf, it will reach 17.9 ft/s just before it hits the ground. Solving this problem also gave us a very useful equation: v_2 = sqrt(2gh_1). This equation can be applied for any object on any planet. However, it will not apply for any height. You may have noticed that it only considers the effects of gravity and ignores air resistance. Thus, this equation wouldn't work for objects like a piece of paper that are strongly affected by air resistance. It also wouldn't work for an object that was high enough up that it reached terminal velocity before it hit the ground. It works for ideal situations in which things like air resistance isn't an issue. However, there are many situations that are very close to ideal. It can be used to get a very close approximation for objects that are dense enough that air resistance isn't a major factor and are low enough that terminal velocity isn't an issue. Thus, it will be useful for most everyday situations.
It would also be pretty easy to solve if the book had an initial velocity. Perhaps counterintuitively, it doesn't matter whether that initial velocity is up or down because velocity is squared. You can understand this pretty simply. If the book is thrown down so it has initial velocity downward, the book will have a higher velocity when it hits the ground. If the book is thrown upward, then the book will go up for a while and then fall down from a higher height. This equation allows us to see that, so long as the magnitude of the velocity is equal, the direction doesn't matter.
The units for energy are Joules. 1 J = 1 N * m = 1 kg * m^2 / s^2.
I should probably stop there before I run away into work, calculus, kinematics, and a gazillion other things that are related to this concept. Hopefully this made some sense and wasn't too boring or difficult to understand.
