1. **State the problem:** We have a cylindrical tank with radius $r=2.5$ feet being drained at a volume rate of $\frac{dV}{dt} = -0.25$ ft$^3$/sec (negative because volume is decreasing). 2. **Formula used:** The volume $V$ of a cylinder is given by $$V = \pi r^2 h,$$ where $h$ is the height of the water. 3. **Differentiate with respect to time $t$:** Since $r$ is constant, differentiate both sides: $$\frac{dV}{dt} = \pi r^2 \frac{dh}{dt}.$$ 4. **Solve for $\frac{dh}{dt}$:** $$\frac{dh}{dt} = \frac{1}{\pi r^2} \frac{dV}{dt}.$$ 5. **Substitute known values:** $$r = 2.5, \quad \frac{dV}{dt} = -0.25,$$ so $$\frac{dh}{dt} = \frac{-0.25}{\pi (2.5)^2} = \frac{-0.25}{\pi \times 6.25} = \frac{-0.25}{19.634954} \approx -0.0127 \text{ ft/sec}.$$ 6. **Interpretation:** The height of the water is decreasing at approximately $0.0127$ feet per second.