**Non-Calculus
Derivation for the Energy-Equation for a book falling off a Table**

Because the acceleration is constant, when we drop the book, it begins to fall (downwards) and its velocity increases (in the negative direction, if up is defined as the positive direction) in a simple linear way with time:

v = - g t

In other words, the
velocity's magnitude increases proportionally with time t, and the
proportionality constant is simply "g", the acceleration of gravity,
by *definition*. For example, given that the acceleration g is about 10
meters per second squared, then after 1 second, the velocity is 10 meters per
second, after 2 seconds its 20 meters per second, after 3 seconds its 30 meter
per second, and so on.

Thus you can see that
this simple formula gives the *instantaneous*, or *momentary* value of
the book's velocity at any given time t you choose.

Now, most people are familiar with the notion that "distance equals velocity times time":

Distance = Velocity x Time

So, you might suppose that we could write:

h = h_{initial}
+ v t

Here, we are using
"h" for distance, which in this case refers to the "height",
or distance of the book above the ground. The symbol "h_{initial}"
represents the initial height of the book (which is a constant).

This is *almost *correct,
but strictly speaking, this formula only holds if the velocity in question is
either *constant* (which is not true in our case), or alternatively, its
true if we use the *average* velocity v_{avg } that the book
had during any particular time *interval* from t=0 till time t=t_{final}.
Hence we should really write:

h = h_{initial}
+ v_{avg }t_{final},

where v_{avg }
is the average velocity that corresponds to the particular value of t_{final }used.

So how do we calculate
v_{avg}?_{ }This is the crucial step! (Its exactly here that we
really avoid the tools of calculus).

Because the *instantaneous*
velocity is given by v = g t (from above), a nice linear relationship, it
follows that the *average* velocity v_{avg
}for the interval ending at t=t_{final } must equal to the
velocity at the *mid-point* of the time interval (at t = 1/2 t_{final }):

v_{avg }= - g
x ( 1/2 t_{final }) = - 1/2 g t_{final }

Putting this back into our formula for h above, we obtain:

h = h_{initial}
+ v_{avg }t_{final} = h_{initial} + (- 1/2 g t_{final
}) x t_{final}

or,

h = h_{initial}
- 1/2 g (t_{final})^{2}

So now we have figured out some really
significant - that the books height decreases according to the *square* of
the time. This is exactly what always occurs to a position variable for
something under *constant* acceleration.

Now you can actually begin to
understand the energy equation above, that is, ** Total Energy
= m g h + 1/2 m v ^{2 }= **

Now, to actually get the energy
equation, we just rearrange our new formula (h
= h_{initial} - 1/2 g (t_{final})^{2}),
and also a formula relating t_{final
}and v, (that is, v = g t_{final
}) to eliminate the explicit appearance of
t_{final }in the
equation.

First, we move the
term containing t_{final }to the left side (for those who want rigor,
this is accomplished by *adding* 1/2 g (t_{final})^{2} to
both sides). We obtain:

h + 1/2 g (t_{final})^{2
}= h_{initial}

Now note that we have

v = - g t_{final }

(This is just the
formula for the instantaneous velocity, as above, at the particular time t_{final
.})

We can rearrange this to read:

t_{final }=
- v_{avg }/ g.

Substitute this
expression into h + 1/2 g (t_{final})^{2 }= h_{initial},
we obtain:

h + 1/2 g (- v_{avg
}/ g)^{2 }= h_{initial}

We're almost there!

The second term simplifies:

1/2 g ( - v_{avg }/
g)^{2 } = 1/2 (v_{avg})^{2
}/ g

(Note that the -1 dissappears because it gets squared)

Now our equation now reads:

h + 1 / 2 (v_{avg})^{2
}/ g = h_{initial}

Now we simply multiply both sides through by mg (just the mass times acceleration - a constant), to get

m g h + 1/2 m v^{2
}= m g h_{initial}

This is our hard-earned energy equation! Note that the right hand side is a constant - its the initial potential energy (which equals the total potential energy).