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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.

Emilio Pisanty
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Kiran Kharel
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  • Related: https://physics.stackexchange.com/q/27847/2451 – Qmechanic Apr 24 '18 at 20:10
  • Why are you questioning the 2? Why aren't you questioning the $g$? It's not always $g$. – Bill N Apr 24 '18 at 20:24
  • Do you question the $\frac{1}{2}$ in $y=y_0+v_{0y}t+\frac{1}{2}a t^2 $? – Bill N Apr 24 '18 at 20:33
  • Although you want to exclude a calculus explanation, I am going to give it to you anyway: $$\frac{dv}{dt} = \frac{dv}{dy}\frac{dy}{dt} = v\frac{dv}{dy}=\frac{1}{2}\frac{dv^2}{dy} = g$$ Now integrate for $v^2$ in the last equality. – nluigi Apr 25 '18 at 07:02

2 Answers2

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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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