Water drips from a shower nozzle onto the floor 198 cm below. The drops fall at regular intervals, with the first drop hitting the floor as the fourth drop begins to fall. When the first drop hits the floor, the second and third drops are at the following distances below the nozzle: (a) second drop and (b) third drop.

 To solve this problem, we will use the equations of motion under gravity. When the drops are in free fall, their motion can be described by the formula for distance traveled under constant acceleration:

\[ s = ut + \frac{1}{2}at^2 \]

where: - \( s \) is the distance traveled, - \( u \) is the initial velocity (which we assume to be 0 since the drop starts from rest), - \( t \) is the time, - \( a \) is the acceleration due to gravity (which we will take as \( 9.8 \, m/s^2 \)).

From the information given, we know the time interval between the drops is constant, so we can find the time it took for the first drop to fall 1.98 m as follows:

For the first drop, \( s = 1.98 \, m \), \( u = 0 \), \( a = 9.8 \, m/s^2 \): \[ 1.98 = 0 \cdot t + \frac{1}{2}(9.8)t^2 \] \[ 1.98 = 4.9t^2 \] \[ t^2 = \frac{1.98}{4.9} \] \[ t \approx \sqrt{0.404} \] \[ t \approx 0.636 \, s \]

Now we know the time it takes for a drop to fall 1.98 m. Since the drops are released at regular intervals, the second drop has been falling for one interval less, and the third drop for two intervals less. Let's call the interval time between drops \( \Delta t \).

1. For the second drop: \( t \) for second drop = \( t \) for first drop - \( \Delta t \) \[ s = \frac{1}{2}(9.8)(t - \Delta t)^2 \]

2. For the third drop: \( t \) for third drop = \( t \) for first drop - \( 2\Delta t \) \[ s = \frac{1}{2}(9.8)(t - 2\Delta t)^2 \]

We are told that the fourth drop starts to fall as the first drop hits the ground, meaning \( t \) for the first drop is equivalent to the time for 3 intervals: \[ t = 3\Delta t \] \[ \Delta t = \frac{t}{3} \] \[ \Delta t = \frac{0.636}{3} \approx 0.212 \, s \]

Now we can find the distances of the second and third drops from the nozzle when the first drop hits the floor:

(a) For the second drop: \[ s = \frac{1}{2}(9.8)(0.636 - 0.212)^2 \] \[ s = \frac{1}{2}(9.8)(0.424)^2 \] \[ s = 4.9(0.179) \] \[ s \approx 0.876 \, m \]

(b) For the third drop: \[ s = \frac{1}{2}(9.8)(0.636 - 2 \cdot 0.212)^2 \] \[ s = \frac{1}{2}(9.8)(0.212)^2 \] \[ s = 4.9(0.045) \] \[ s \approx 0.220 \, m \]

So, when the first drop hits the floor, the second drop is approximately 0.876 meters below the nozzle, and the third drop is approximately 0.220 meters below the nozzle.