Why is it $q^2$ for the individual count in hardy weinberg?
First of, let me correct your equation: $q = \sqrt{\frac{1}{100}} = \frac{1}{10} = 0.1 ≠ 10 $.

From allele frequency to genotype frequency

Imagine you were to randomly sample an allele from a population of allele where the allele A is present at frequency $q$. What is the probability that you draw allele A? Answer: $P(A) = q$. Now, put this allele back in the pool and imagine you have to draw two alleles. What is the probability that the two alleles are A. Well it is the probability that the first allele is A times the probability that the second allele is A, it is $P(A) \cdot P(A) = q \cdot q = q^2$.

From genotype frequency to allele frequency

Therefore, under Hardy-Weinberg conditions, if the allele A is at frequency $q$, then the homozygote genotype AA is at frequency $q \cdot q = q^2$. Now let's denote the frequency of the genotype AA by $f_{AA}$. You know that $f_{AA} = q^2$ from the above though experiment. If you take the square root of both sides, you get $\sqrt{f_{AA}} = \sqrt{q^2} = q$. In words, the frequency of the allele A is the square root of the frequency of the genotype AA.

Recessivity and dominance

Btw, you'll note that those calculations tell nothing about the patterns of dominance/recessivity of alleles (Hardy-Weinberg principle assumes no selection anyway).