A tank containing water is in the form of a cone with vertex C. The axis is vertical and the semi-vertical angle is 60°, as shown in the diagram. At time t = 0, the tank is full and the depth of water is H. At this instant, a tap at C is opened and water begins to flow out. The volume of water in the tank decreases at a rate proportional to \(\sqrt{h}\), where h is the depth of water at time t. The tank becomes empty when \(t = 60\).

(i) Show that h and t satisfy a differential equation of the form \(\frac{dh}{dt} = -Ah^{-\frac{3}{2}}\), where A is a positive constant.

(ii) Solve the differential equation given in part (i) and obtain an expression for t in terms of h and H.

(iii) Find the time at which the depth reaches \(\frac{1}{2}H\).

[The volume V of a cone of vertical height h and base radius r is given by \(V = \frac{1}{3} \pi r^2 h\).]