What is the location of the second and third water drops in a falling shower?

[rocomath]

Water drips from the nozzle of a shower onto the floor 200cm below. The drops fall at regular equal intervals of time, the first drop striking the floor at the instant the fourth begins to fall. Find the locations of the second and third drops when the first strikes the floor.

First I found the total time it takes for 1 drop to hit the bottom of the floor with equation (1): $$y-y_0=v_0t-\frac 1 2 gt^2$$

$$\Delta y =-200cm$$
$$v_0=0m/s$$

Solved for t: $$t=\sqrt \frac{-2y}{g}=\sqrt \frac{-2 \times -0.200m}{9.81 m/s}}\approx .202s$$

From there, I split the time into 4 equal intervals, and used those times I got and plugged it back into equation (1). But this method isn't correct, so I'm not sure where to go from here.

 

[hage567]

From there, I split the time into 4 equal intervals

Why do you think it should be split into 4 intervals?

 

[rocomath]

Why do you think it should be split into 4 intervals?

Well the problem states "The drops fall at regular equal intervals of time" and so that's what I went by.

 

[t!m]

I think hage is suggesting, why 4 and maybe not ... another number of intervals? Think about the location of the droplets when the first hits the floor.

 

[hage567]

Maybe draw a diagram with the drops on it. The question states that when the first one hits the ground, the fourth is just about to start falling. So if there are two other drops in mid-air, how many intervals does that make?

 

[rocomath]

There would be 3 intervals.

1st droplet from the ground to the 2nd (1)
2nd droplet to the 3rd, also in motion (2)
3rd droplet from the 4th about to fall (3)

Works :-] Thanks hage and t!m.