We can use the formula for the volume of a cone to solve this problem. Let r be the radius of the cone and h be the height of the water. Then the volume of the water in the cone is given by:
V = (1/3)πr^2h
We are given that the cone is initially filled with water to a depth of 5 cm. Since the vertex is downwards, this means that the height of the water is h = 10 cm - 5 cm = 5 cm. We are also given that adding 720 cubic cm of water raises the level to 10 cm. Let V1 be the volume of water in the cone before the additional water is added, and let V2 be the volume of water in the cone after the additional water is added. Then we have:
V1 = (1/3)πr^2(5)
V2 = (1/3)πr^2(10)
Adding 720 cubic cm of water increases the volume of the water in the cone by:
V2 - V1 = (1/3)πr^2(10) - (1/3)πr^2(5) = (1/3)πr^2(5)
We are asked to find the quantity of water that must be added in order to raise the level to 15 cm. Let V3 be the volume of water in the cone after this additional water is added. Then we have:
V3 = (1/3)πr^2(15)
We want to find the volume of water that needs to be added to go from V2 to V3. This is given by:
V3 - V2 = (1/3)πr^2(15) - (1/3)πr^2(10) = (1/3)πr^2(5)
But we know that adding 720 cubic cm of water increased the volume of the water in the cone by the same amount. So we have:
(1/3)πr^2(5) = 720
Solving for r, we get:
r^2 = (720*3)/(5*π) = 432/π
r ≈ 9.28 cm
Finally, we can find the volume of water that needs to be added to go from V2 to V3:
V3 - V2 = (1/3)π(9.28