
Torricelli's law (3)
Suppose that an initially full hemispherical water tank of radius 1 m has its flat side as its bottom. It has a bottom hole of radius 1 cm. If this bottom hole is opened at 1 P.M., when will the tank be empty?
My working:
$\displaystyle A(y)=\pi r^2=\pi(1y^2)$
$\displaystyle \pi(1y^2) \frac{dy}{dt}=\pi\left(\frac{1}{100}\right)^2 \sqrt{2*9.8y}$
$\displaystyle \int(y^{1/2}y^{3/2})dy=\int{\frac{\sqrt{19.6}}{10^4} dt$
$\displaystyle 2y^{1/2}\frac{2}{5}y^{5/2}=\frac{\sqrt{19.6}}{10^4} t+C$
$\displaystyle y(0)=1$
$\displaystyle C=2\frac{2}{5}=\frac{8}{5}$
$\displaystyle 2y^{1/2}\frac{2}{5}y^{5/2}=\frac{\sqrt{19.6}}{10^4} t+\frac{8}{5}$
When $\displaystyle y=0$,
$\displaystyle t=\frac{80000}{5*\sqrt{19.6}}\approx 3614\ s$
The tank is empty about 14 seconds after 2:00 P.M.