# Calculating kinetic energy?

Would this be a valid equation to calculate kinetic energy created from a drop from a height:

$$E_{kinetic} ~=~ v_{vertical}tmg$$

Velocity multiplied by time gives distance. Distance multiplied by gravitational force acting on it provides kinetic energy. Would this equation be valid?

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The answer to the question(v1) is Yes if $v_{vertical}=\frac{gt+0}{2}$ denotes the average velocity during the drop. –  Qmechanic May 23 '12 at 15:39
Although, what you might be getting at would be that $E_{kinetic}=\int v_{vertical}mg \cdot dt$, which is identically $E_{kinetic}=mgh$. If you don't know calculus, the equation $dE_{kinetic}=v_{vertical}mg \cdot dt$ might make more sense. Because v changes over time, you can only say that "h=v*dt" over very small values of "dt", so this gives you a very small portion of energy, and when you add all the tiny portions of energy together you get the correct result. –  NeuroFuzzy Sep 8 '12 at 6:55

No.

You are correct that the kinetic energy is equal to the change in the potential energy, $mgh$, where $h$ is the distance fallen, but because the object is accelerating $h$ is not simply velocity times time. If the object starts at rest then (ignoring air resistance):

$$h = \frac{1}{2} g t^2$$

so substituting this into $mgh$ gives:

$$E_{kinetic} = \frac{1}{2} m g^2 t^2$$

Note that $gt$ is just the velocity at time $t$, so this expression is the same as:

$$E_{kinetic} = \frac{1}{2} m v^2$$

which may look familiar :-) Note however that the velocity $v$ is a function of time.

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Thank You John! –  user8791 May 23 '12 at 15:40
For a body of mass m in a uniform specific-force field $g$:
$$E_{kinetic} = \frac{1}{2} m g^2 t^2$$
For two mutually interacting bodies of masses M and m: $$E_{kinetic} = \frac{G(M+m)}{r_1}-\frac{G(M+m)}{r_0}$$
where $r_0$ and $r_1$ are the initial and final separations of the two bodies.