Assuming that in case of motion of blunt bodies in air acrodynamic drag depends on effective area a of the body, the speed of body relative to air v and density of air `sigma`, show by method of dimensions:
`D=K sigma Av^(2)`

Explanation

Dimensions of 'D' that of force `=[MLT^(-2)]`
Dimension of `sigma=[ML^(-3)]`
Dimension of `A=[L^(2)]`
Dimension of `v=[LT^(-1)]`
k is dimensions less
Let, `D=K (sigma)^(a) (A)^(b) (v)^(c)`
`:. [MLT^(-2)]=[ML^(-3)]^(a)[L^(2)]^(b)[LT^(-1)]^(c)`
`:.` from LHS and RHS
`1=a`
`1=-3a+2b+c`
`-2=-c`
So, `a=1, c=2` and `b=1`
`:. D=K alpha Av^(2)`