1. **State the problem:** We are given the time $t$ in seconds for an object dropped from height $s$ meters to reach the ground, with the formula $$t = \sqrt{\frac{s}{5}}.$$ We need to find the rate of change of time with respect to height, i.e., $\frac{dt}{ds}$, when $s = 125$ meters. 2. **Recall the formula and rules:** The derivative of $t$ with respect to $s$ is found using the chain rule. Since $t = \sqrt{\frac{s}{5}} = \left(\frac{s}{5}\right)^{\frac{1}{2}}$, we use the power rule: $$\frac{dt}{ds} = \frac{1}{2} \left(\frac{s}{5}\right)^{-\frac{1}{2}} \cdot \frac{d}{ds}\left(\frac{s}{5}\right).$$ 3. **Calculate the derivative inside:** $$\frac{d}{ds}\left(\frac{s}{5}\right) = \frac{1}{5}.$$ 4. **Substitute back:** $$\frac{dt}{ds} = \frac{1}{2} \left(\frac{s}{5}\right)^{-\frac{1}{2}} \cdot \frac{1}{5} = \frac{1}{10} \left(\frac{s}{5}\right)^{-\frac{1}{2}}.$$ 5. **Rewrite the negative exponent:** $$\frac{dt}{ds} = \frac{1}{10} \cdot \frac{1}{\sqrt{\frac{s}{5}}} = \frac{1}{10} \cdot \frac{1}{\frac{\sqrt{s}}{\sqrt{5}}} = \frac{1}{10} \cdot \frac{\sqrt{5}}{\sqrt{s}} = \frac{\sqrt{5}}{10 \sqrt{s}}.$$ 6. **Evaluate at $s=125$:** $$\frac{dt}{ds} \bigg|_{s=125} = \frac{\sqrt{5}}{10 \sqrt{125}}.$$ 7. **Simplify $\sqrt{125}$:** $$\sqrt{125} = \sqrt{25 \times 5} = 5 \sqrt{5}.$$ 8. **Substitute and simplify:** $$\frac{dt}{ds} = \frac{\sqrt{5}}{10 \times 5 \sqrt{5}} = \frac{\sqrt{5}}{50 \sqrt{5}}.$$ 9. **Cancel $\sqrt{5}$:** $$\frac{dt}{ds} = \frac{\cancel{\sqrt{5}}}{50 \cancel{\sqrt{5}}} = \frac{1}{50}.$$ **Final answer:** The rate of change of time with respect to height when the object is 125 m above the ground is $$\boxed{\frac{1}{50}} \text{ seconds per meter}.$$