I'm working through The Cartoon Guide to Physics and I can't figure out how an equation was rearranged for the constant acceleration formula $d = {1\over2}gt^2$. The object will fall 4 ft and the book gave $t = {\sqrt{1 sec^2/4}} = {1\over 2} sec$ as the solution.
My attempt was (edited thanks to N. F. Taussig):
$$\begin{align} 4ft &= {1\over 2}(32ft/sec^2)t^2 \\ {4ft \over t^2} &= {1\over 2}(32ft/sec^2) \\ {1\over t^2} &= {{{1\over 2}(32ft/sec^2)} \over 4ft} \\ {1\over t^2} &= {{16ft/sec^2} \over 4ft} \\ {1\over t^2} &= {{4 sec^2 }} \\ t^2 &= {1 \over {4 sec^2}} \\ \sqrt{t^2} &= {\sqrt{1} \over \sqrt{4 sec^2}} \\ t &= { 1 \over 2 }sec \end{align}$$
I don't see where their $t = {\sqrt{1 sec^2/4}}$ came from. Hopefully it's not too much to ask not only where I'm going wrong but what algebra rules I need to study to make this come more naturally in the future.
You know that
$$d = \frac{1}{2}gt^2$$
Solve for $t$.
\begin{align*} d & = \frac{1}{2}gt^2\\ 2d & = gt^2\\ \frac{2d}{g} & = t^2\\ \sqrt{\frac{2d}{g}} & = t \end{align*}
Substitute for $d$ and $g$ to obtain
$$t = \sqrt{\frac{2 \cdot 4~\text{ft}}{32~\frac{\text{ft}}{\text{s}^2}}} = \sqrt{8~\text{ft} \cdot \frac{1}{32}~\frac{\text{s}^2}{\text{ft}}} = \sqrt{\frac{1}{4}~\text{s}^2} = \frac{1}{2}~\text{s}$$