The question asks you to guess a plausible formula for the volume V of the crater produced by a surface explosion with energy E in soil of density ρ with gravitational constant given, as usual by g, using the formula: V = k(E^a x ρ^b x g^c) for some constant k.
We use these units:
E measured in kg. m^2. sec^-2
ρ measured in kg . m^-2
g measured in m. sec^-2
Using this info, find values for a, b and c.
I started by assuming that V would be measured in m^3, and used this along with the units above in the formula.
After combining like terms, I obtained:
m^3 = k[kg^(a+b) x m^(2a-3b+c) x sec^(-2a-2c)]
I got stuck here because although I know I will have to solve the equations simultaneously at some point, I don't know what the next step would be here in terms of finding what each expression would be equal to, if that makes sense.
$ \require{enclose} \def\a{\alpha} \def\b{\beta} \def\g{\gamma} \def\e{\enclose{updiagonalstrike}} \def\LR#1{\left(#1\right)} \def\FLR#1#2{\LR{\frac{#1}{#2}}} $Assuming the density is volumetric and not area$\ell$, the equation to be balanced is $$\eqalign{ V &= E^\a \rho^\b g^\g \\ \LR{m^3} &= \FLR{kg\cdot m^2}{s^2}^\a\:\FLR{kg}{m^3}^\b\:\FLR{m}{s^2}^\g \\ }$$ To cancel units of $(kg,\,s)$ which only appear on the RHS, you must set $\;\b=\g=-\a$ $$\eqalign{ \LR{m^3} &= \FLR{\e{kg}\cdot m^2}{\e{s^2}}^\a \FLR{m^3}{\e{kg}}^\a \FLR{\e{s^2}}{m}^\a = \LR{m^4}^\a \\ }$$ Solving for the remaining exponent yields $$ \a = \frac34 \quad\implies\quad \b=\g=-\frac34 $$ This leaves $k$ as a dimensionless constant of proportionality.