Answer:
The height of the water is [tex]60.5\ ft[/tex]
Step-by-step explanation:
step 1
Find the volume of the tank
The volume of the inverted right circular cone is equal to
[tex]V=\frac{1}{3}\pi R^{2} H[/tex]
we have
[tex]R=16\ ft[/tex]
[tex]H=96\ ft[/tex]
substitute
[tex]V=\frac{1}{3}\pi (16)^{2} (96)[/tex]
[tex]V=8,192\pi\ ft^{3}[/tex]
step 2
Find the 25% of the tank’s capacity
[tex]V=(0.25)*8,192\pi=2,048\pi\ ft^{3}[/tex]
step 3
Find the height, of the water in the tank
Let
h ----> the height of the water
we know that
If two figures are similar, then the ratio of its corresponding sides is proportional
[tex]\frac{R}{H}=\frac{r}{h}[/tex]
substitute
[tex]\frac{16}{96}=\frac{r}{h}\\ \\r= \frac{h}{6}[/tex]
where
r is the radius of the smaller cone of the figure
h is the height of the smaller cone of the figure
R is the radius of the circular base of tank
H is the height of the tank
we have
[tex]V=2,048\pi\ ft^{3}[/tex] -----> volume of the smaller cone
substitute
[tex]2,048\pi=\frac{1}{3}\pi (\frac{h}{6})^{2}h[/tex]
Simplify
[tex]221,184=h^{3}[/tex]
[tex]h=60.5\ ft[/tex]
