## Fun with dimensional analysis 2 – Projectile range

The problem facing every high school and introductory college physics student is that of a projectile fired at an angle ignoring air resistance. It’s fairly simple to derive from the first principles differential equations (with initial velocity $v_0$, angle $\theta$, and vertical acceleration $\displaystyle a_y = -g$. In this analysis, we wish to know what the cannoneers of old wanted to know: What is the range, or horizontal distance traveled before the cannon ball strikes the ground? We’ll call the range R, though in the following diagram it shows up as x.

We wish to know the range R, and we know $v_0$ and $\theta$. Again, $\theta$ is dimensionless so dimensional analysis will tell us nothing about it. I don’t include the ball’s mass since we know that it will have no effect, ie. if we include it in the dimensional analysis it will have power 0 just like in the pendulum’s case. What’s the relevant physics we need to consider? There’s no air resistance, but there’s certainly gravity so we include g. Then we have

$R \displaystyle \sim v_0^\alpha g^\beta \Rightarrow \lbrack R \rbrack = \lbrack v_0 \rbrack^\alpha \lbrack g \rbrack^\beta \Rightarrow L = L^\alpha T^{-\alpha} L^\beta T^{-2\beta}$

$\displaystyle \Rightarrow 1 = \alpha + \beta \mbox{ and } 0 = -\alpha - 2\beta$

$\displaystyle \Rightarrow \beta = -1 \mbox{ and } \alpha = 2 \mbox{ so}$

$\displaystyle R \sim \frac{v_0^2}{g}$

We could leave it at that, or we could consider the limiting or easy cases of $\theta$ to guess the angle dependence of the range. For $\displaystyle \theta = 0$, the ball leaves the cannon and then immediately strikes the ground so $\displaystyle R = 0$. For $\displaystyle \theta = \frac{\pi}{2}$ the ball flies straight up and then returns to the origin, so we again have $\displaystyle R = 0$. The first condition is satisfied by including, in the equation for range, a $\displaystyle \sin( \theta)$ term and the second is correspondingly satisfied with a $\displaystyle \cos( \theta)$ term. Therefore,

$\displaystyle R \sim \frac{v_0^2}{g} \cos( \theta) \sin( \theta)$.

The actual formula is

$\displaystyle R = 2 \frac{v_0^2}{g} \cos( \theta) \sin( \theta)$.