Growing Tree

Let's first solve for the growth constant $k$ : at $h(0) = 1, \frac{dh}{dt} = 0.4$ m/yr = $0.0333$ m/mo. Solving for $k$ gives:

$0.0333 = k \cdot (9 - 1)^{1/3} \Rightarrow k = 0.0167 = \frac{1}{60}$

The tree's maximum height occurs when $\frac{dh}{dt} = 0 \Rightarrow h = 9.$ Solving the original differential equation (via Wolfram) produces:

$h(t) = \frac{t}{270} \cdot \sqrt{36 - \frac{t}{10}} - \frac{2}{3} \cdot \sqrt{144 - \frac{2t}{5}} + 9$

The required time T to reach this maximum height $h(T) = 9$ computes to:

$9 = \frac{T}{270} \cdot \sqrt{36 - \frac{T}{10}} - \frac{2}{3} \cdot \sqrt{144 - \frac{2T}{5}} + 9$ ;

or $\frac{T}{270} \cdot \sqrt{36 - \frac{T}{10}} - \frac{2}{3} \cdot \sqrt{144 - \frac{2T}{5}} = 0$ ;

or $\frac{360T^2 - T^3}{729000} = \frac{2880 - 8T}{45}$ ;

or $45(T+360)(T-360)^2 = 0$ ;

$T = 360$ months = $\boxed{30}$ years.