## Statement

A broken faucet lets out water drops every 1/4 of a second. If the faucet is just 3 meters above the ground, and if one drop falls in this very instant, determine what is the position of the drops still in the air at this moment.

## Solution

**Data**

H = 3 m

g = 9.8 m/s^{2}

Time that each drop has been in the air

t_{0}=0 s, t_{1}=0.25 s, t_{2}=0.5 s, t_{3}=0.75 s, t_{4}=1 sg, etc...

y_{0}, y_{1}, y_{2}, y_{3}, ... ?

**Resolution**

Let us call y_{0}, y_{1}, y_{2}... the positions of each drop in the air, starting with the one still exiting the faucet as it is shown in the figure. In the same way, let us call t_{0}, t_{1}, t_{2}, etc... the time each of these drops has been in the air.

Applying the formula of the position in fall free motion, for each drop, we get that:

$$y=H-\frac{1}{2}\xb7g\xb7{t}^{2}\phantom{\rule{0ex}{0ex}}{y}_{0}=3-\frac{1}{2}\xb79.8\xb7{\left(0\right)}^{2}=3m\phantom{\rule{0ex}{0ex}}{y}_{1}=3-\frac{1}{2}\xb79.8\xb7{\left(0.25\right)}^{2}=2.69m\phantom{\rule{0ex}{0ex}}{y}_{2}=3-\frac{1}{2}\xb79.8\xb7{\left(0.5\right)}^{2}=1.77m\phantom{\rule{0ex}{0ex}}{y}_{3}=3-\frac{1}{2}\xb79.8\xb7{\left(0.75\right)}^{2}=0.24m\phantom{\rule{0ex}{0ex}}{y}_{4}=3-\frac{1}{2}\xb79.8\xb7{\left(1\right)}^{2}=-1.9m\phantom{\rule{0ex}{0ex}}....\phantom{\rule{0ex}{0ex}}$$

As you can see, the position of drop y_{4} is a negative number, so, if the ground is y=0, then the drop reached the ground at some prior time and therefore, it is not in the air at this moment. From that we reason that there are only 4 drops still falling.