Drops in fall free

Data

H = 3 m
g = 9.8 m/s²

Time that each drop has been in the air
t₀=0 s, t₁=0.25 s, t₂=0.5 s, t₃=0.75 s, t₄=1 sg, etc... 

y₀, y₁, y₂, y₃, ... ?

Resolution

Let us call y₀, y₁, y₂... 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 t₀, t₁, t₂, 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:

$$\begin{gathered} y=H-\frac{1}{2}·g·t^2 \\ {y}_0=3-\frac{1}{2}·9.8·{\left(0\right)}^2 = 3 m \\ {y}_1=3-\frac{1}{2}·9.8·{\left(0.25\right)}^2 =2.69 m \\ {y}_2=3-\frac{1}{2}·9.8·{\left(0.5\right)}^2 =1.77 m \\ {y}_3=3-\frac{1}{2}·9.8·{\left(0.75\right)}^2 =0.24 m \\ {y}_4=3-\frac{1}{2}·9.8·{\left(1\right)}^2=-1.9 m \\ .... \end{gathered}$$

As you can see, the position of drop y₄ 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.