Answer
(a) $t=\frac{\sqrt {55}}{4}\approx1.85\,s$.
(b) $0.41\text{ s}$ sooner.
Work Step by Step
(a) $h=55\text{ ft}$
$t=\sqrt{\frac{h}{16}}=\sqrt {\frac{55}{16}}=\frac{\sqrt {55}}{\sqrt {16}}=\frac{\sqrt {55}}{4}$
$t=\frac{\sqrt {55}}{4}$ or about $1.85\,s$.
(b) $h=(55-22)\text{ ft}=33\text{ ft}$
$t=\sqrt{\frac{33}{16}}=\frac{\sqrt {33}}{\sqrt {16}}=\frac{\sqrt {33}}{4}$
$t=\frac{\sqrt {33}}{4}$ or about $1.44\text{ sec}$
$\Delta t=(1.85-1.44)\text{ sec}=0.41\text{ sec}$
The earring hits the ground $0.41\text{ sec}$ sooner.
