4- 195 Two drops of water fall successively from a steep cliff that is 300 meters high. When the first drop has fallen 0.001 millimeters, the second drop begins to fall. What is the distance between the two drops when the first drop reaches the bottom of the cliff (the answer should be accurate to 0.1 millimeters; air resistance is negligible)?
[Solution] Let $d$ be the distance between the two water droplets when the second droplet starts to fall, and $s_{1}$ and $s_{2}$ be the distances traveled by the first and second droplets, respectively, when the first droplet reaches the foot of the mountain. If the distances $d, s_{1}$, and $s_{2}$ are traveled in times $t, t_{1}$, and $t_{1}-t$, respectively, then
$$d=\frac{g t^{2}}{2}, s_{1}=\frac{g t_{1}^{2}}{2}, s_{2}=\frac{g\left(t_{1}-t\right)^{2}}{2}$$

Therefore, $s_{2}=\frac{g}{2}\left(\sqrt{\frac{2 s_{1}}{g}}-\sqrt{\frac{2 d}{g}}\right)^{2}=s_{1}-2 \sqrt{s_{1} d}+d$
which means $s_{1}-s_{2}=2 \sqrt{s_{1} d}-d$.
In this problem, $d=0.001$ millimeters, $s_{1}=300000$ millimeters, so $s_{1}-s_{2}=34.6$ millimeters