We have a tank of conical shape (as an inverted cone, with the apex below and the basis up) to fill it with rainwater. The entire tank measures $$5$$ metres high. Define the radius of the cone so, when the water rises the half of its height, it contains $$1000$$ L of water.
Development:
First, we have to find the dimensions of the part that contains water, which is the half a cone. So, on one hand, $$h' =\dfrac{5}{2}=2,5$$.
On the other hand, using the information of the volume: $$$V_{water}=1000L\cdot\dfrac{1\ m^3}{1000L}=1\ m^3$$$ $$$V_{water}=\dfrac{1}{3}\cdot h_{water}\cdot\pi\cdot(r_{water})^2$$$ $$$r_{water}=\sqrt{\dfrac{3\cdot V_{water}}{\pi\cdot h_{water}}}=0,618 \ m$$$
Finally, we use the proportions (because there is a direct relation between $$h$$ and $$r$$) $$$\dfrac{r_{water}}{r_{tank}}=\dfrac{h_{water}}{h_{tank}} $$$ $$$r_{tank}=r_{water}\cdot\dfrac{h_{tank}}{h_{water}}=1,236 \ m^3$$$
Solution:
$$r_{tank}=1,236 \ m^3$$