## Fun with dimensional analysis 3 – Vertical throw

A ball is thrown into the air with initial velocity $\displaystyle v_0$ and we wish to know how long it takes to return to the hand neglecting air resistance. The relevant physics is just the gravitational acceleration $\displaystyle g$. Thus we have

$\displaystyle t \sim v_0^\alpha g^\beta \\ \Rightarrow \lbrack t \rbrack =\lbrack v_0 \rbrack^\alpha \lbrack g \rbrack^\beta \Rightarrow T = L^\alpha T^{-\alpha} L^\beta T^{-2\beta} \\ \Rightarrow 1 = -\alpha - 2\beta \mbox{ and } 0 = \alpha + \beta \\ \Rightarrow \beta = -1 \mbox{ and } \alpha = 1 \\ \therefore t \sim \frac{v_0}{g}$

This result is fairly trivial to derive via differential equations (the missing constant is 2) but dimensional analysis is such an effective tool that even in this trivial cases I am wont to try it out.

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