How to solve the problem by using separation of variables? A tank in the form of a right circular cone of height $1.6$ meters and radius $1$ meter with its vertex below the base is filled with water. A $1$ centimeter hole at the vertex causes water to leak out. Determine how long it takes for the tank to be empty. Assume that the rate of the water flow out through the hole is given by $0.6 \, \sqrt{2gh}$. Where $h$ is the instantaneous height of the water.




Based on the problem I've came up of the differential equation:
$$\frac{π}{1.6^2}h^2\frac{dh}{dt}=-0.0001\pi[0.6(\sqrt{2gh})]$$
Which i think can be solved by seperation of variables, my current weakness in solving DE.
Thank you.
We have: $h^2\dfrac{dh}{dt} = -0.0001\cdot (1.6)^2\cdot 0.6\cdot \sqrt{2g}\cdot \sqrt{h}= C\sqrt{h}\to h^{1.5}dh = Cdt$. You can continue at this point...
Notice, $$\frac{π}{1.6^2}h^2\frac{dh}{dt}=-0.0001π[0.6(\sqrt{2gh})]$$

$$\frac{1}{1.6^2}h^{3/2}dh=-0.0001[0.6(\sqrt{2g})]dt$$
$$\frac{10^{5}}{15.36\sqrt{2g}}\int h^{3/2}dh=-\int dt$$

Now, if $t$ is the time to empty the  tank from height $h=1.6 \ m$ to $ h=0$ then applying proper limits, we get
$$\frac{10^{5}}{15.36\sqrt{2g}}\int_{1.6}^{0} h^{3/2}dh=-\int_{0}^{t} dt$$
$$\frac{10^{5}}{15.36\sqrt{2g}}\left[\frac{h^{5/2}}{\frac{5}{2}}\right]_{1.6}^{0} =-[t]_{0}^{t}$$
$$\frac{2\times 10^{5}}{76.8\sqrt{2g}}\left[0-(1.6)^{5/2}\right]_{1.6}^{0} =-t$$ $$t=\frac{2\times 10^{5}\times (1.6)^{5/2}}{76.8\sqrt{2\times 9.81}}$$
$$\approx 1903.79\ sec\approx 31.73 \ minutes$$