Water drips from a shower head (the sprayer at the top of the shower) and falls onto the floor 2.3 m below. The droplets are falling at regular intervals (equal amounts of time between drops), with the first drop hitting the floor at the instant that the fourth drop starts to fall. When the first drop hits the floor (as the fourth drop is dripping from the shower head), how far below the shower head is the third drop? [The third drop is the one that fell right before the fourth drop.]

0.767m

Explanation:

We are given that the time interval between each droplet is equal.

We are also given that the fourth drop is just dripping from the shower when the first hits the floor.

If they fall at the same time interval and we know that the distance between the shower head and floor are the same, they must therefore fall at the same velocity.

The distance between each drop has to be the same given that they fall at equal time intervals.

Let this distance be x.

We can then partition the entire height of the system into three parts (as shown in the diagram).

Hence, we can say that:

x + x + x = 2.3m

3x = 2.3m

=> x = 2.3/3 = 0.767m

Therefore, at the time the first drop hits the floor, the third drop is only 0.767 m below the shower head.

it is a negative acceleration