Drops fall from the mouth of a vertical pipe one after

Drops fall from the mouth of a vertical pipe one after another at intervals of $\tau = 0.1\ \text{s}$, each starting from rest. Find the difference in the velocities of the first and second drops one second after the first leaves.

Step 1: Velocity of a single drop

A drop that has been falling for a time $\theta$ has velocity

\[ v = g\theta. \]

Step 2: How long each has been falling

Taking $t = 0$ as the release of the first drop, the second leaves at $t = \tau$. At $t = 1\ \text{s}$: \[\theta_{1} = 1\ \text{s}, \qquad \theta_{2} = 1-\tau = 0.9\ \text{s}.\]

Step 3: The difference

\[ \Delta v = g\theta_{1}-g\theta_{2} = g\bigl(\theta_{1}-\theta_{2}\bigr) = g\tau. \]

\[ \Delta v = 9.81\times0.1 = 0.981\ \frac{\text{m}}{\text{s}}. \]

Step 4: The moment of observation does not matter

The answer $g\tau$ contains no $t$. Both drops gain speed at the same rate $g$, so once the second drop is in flight their velocities differ by a fixed amount — the speed the first had already acquired during the interval $\tau$.

Step 5: Verification

\[t = 1:\ v_{1} = 9.81,\ v_{2} = 8.829,\ \Delta v = 0.981 \;\;\checkmark\]

\[t = 5:\ v_{1} = 49.05,\ v_{2} = 48.069,\ \Delta v = 0.981 \;\;\checkmark\]

\[\text{The gap in position, by contrast, grows: } \Delta s = g\tau t-\frac{g\tau^{2}}{2} \;\;\checkmark\]