In the relationship $$v_y^2=v_{0y}^2+2g(y-y_0),$$ what does the $2g$ mean?
Why do we have gravity being multiplied by 2? Don't say you need to know calculus; I know that. I don't want to know how but why.
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Perhaps you find it more intuitive in another form: $$\frac{1}{2}mv_y^2 = \frac{1}{2}mv_0^2 + mg(y-y_0)$$ This is just a statement of conversion of energy, in this case kinetic and potential energies.
nluigi
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This is a kinematic equation, so it only applies when the acceleration is constant. Under this scenario, the average velocity of a particle over some duration of time $\Delta t$ is
$$v_{\mathrm{avg}} = \frac{1}{2}(v_{f} + v_{o}),$$
where $v_{f}$ is the particle's velocity at the end of $\Delta t$ and $v_{o}$ is its velocity at the beginning of $\Delta t$. The "2" in the fraction is going to carry through the derivation of your equation.
The next step is to write
$$\Delta t = \frac{(v_{f} - v_{o})}{a},$$
and substitute for $v_{\mathrm{avg}}$ and $\Delta t$ in $\Delta x = v_{avg} \Delta t$. Just trade $y$ for $x$ and $g$ for $a$ and work through the algreba.
Rodney Dunning
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