A rectangular reservoir has a horizontal base of area 1000 m². At time \(t = 0\), it is empty and water begins to flow into it at a constant rate of 30 m³s^-1. At the same time, water begins to flow out at a rate proportional to \(\sqrt{h}\), where \(h\) m is the depth of the water at time \(t\) s. When \(h = 1\), \(\frac{dh}{dt} = 0.02\).

(i) Show that \(h\) satisfies the differential equation \(\frac{dh}{dt} = 0.01(3 - \sqrt{h})\).

It is given that, after making the substitution \(x = 3 - \sqrt{h}\), the equation in part (i) becomes \((x - 3) \frac{dx}{dt} = 0.005x\).

(ii) Using the fact that \(x = 3\) when \(t = 0\), solve this differential equation, obtaining an expression for \(t\) in terms of \(x\).

(iii) Find the time at which the depth of water reaches 4 m.