Answer
$ t = \frac{A_0}{A_1}\sqrt{\frac{2h}{g}}$
Work Step by Step
We know that the volume is changing at a rate of $A\frac{dh}{dt}$, and that the volume through the orifice is changing at a rate of $A_1\sqrt{2gh}$. Thus, setting these equal, we find:
$A\frac{dh}{dt}=A_1\sqrt{2gh} \\ \frac{Adh}{\sqrt{2gh}}=A_1dt$
Taking the integral of both sides and simplifying, we find:
$ t = \frac{A_0}{A_1}\sqrt{\frac{2h}{g}}$
