Answer
$\frac{dh}{dt}=\frac{8}{405\pi}feet/min$
Work Step by Step
Step-1: The following data are given to us,
$\frac{dV}{dt} = 10 cm^3/min$
$diameter\approx 3h\implies radius=1.5h$
The relation between volume of the cone and its height is,
$$V = \frac{1}{3}\pi r^2 h$$
Since $r=1.5h$,
$$V = \frac{1}{3}\pi \times (2.25)h^3=0.75\pi h^3$$
Step-2: Differentiate the above equation with respect to $h$,
$$\frac{dV}{dt}=(2.25)\pi h^2 \frac{dh}{dt}$$
$$\implies \frac{dh}{dt}=\frac{dV}{dt} \times \frac{1}{(2.25)\pi h^2}=\frac{10}{(2.25)\pi h^2}$$
Step-3: When $h=15$feet,
$$\frac{dh}{dt}=\frac{10}{(2.25)\pi (15)^2}=\frac{8}{405\pi}feet/min$$