Simplifying $\frac1{gt}\sqrt{g/2h}\,dx$ in free fall equations

Question

The relevant equation: $x(t) = \frac12 gt^2$ , $dx/dt = gt$ , $T=\sqrt{2h/g}$

$dt/T = (dx/gt)\sqrt{g/2h} = 1/(2\sqrt{hx}) dx $

I do not see how $(dx/gt)\sqrt{g/2h}$ turns into $1/(2\sqrt{hx}) dx $

Can someone please show me how this was simplified?

Answer 1

It's just a matter of solving for $t = \sqrt{2x/g}$ and then plug and chug: $$\frac{dx\sqrt{g}}{gt\sqrt{2h}} = \frac{dx\sqrt{g}}{g\sqrt{2x/g}\sqrt{2h}} = \frac{dx}{2\sqrt{xh}}.$$