Water drips from the nozzle of a shower onto the floor below. The drops fall at regular (equal) intervals of time, the first drop striking the floor at the instant the fourth drop begins to fall. When the first drop strikes the floor, how far below the nozzle are the (a) second and (c) third drops?

Given

$\mathrm{g}=-9.8\mathrm{m}/{\mathrm{s}}^2$, in the downward direction.

The distance between nozzle and floor,$∆\mathrm{y}=-2.0\mathrm{m}$.

Four drops are leaving the nozzle simultaneously with an equal interval of time. The time of each second and third drop reaching the ground can be determined by the kinematic equation that is mentioned below.

The kinematic equation that can be used to solve this problem is,

$∆y=V_{i}t+\frac{1}{2}a{t}^2$ (i)

By using the kinematics equation given in equation (i),

Let’s assume t₁ is the time taken by drop to drop on the floor.

$$\begin{gathered} -2m=0-\frac{1}{2}a{t}_1^2 \\ {t}_1=\sqrt{\frac{2∆y}{9.8}} \\ =0.639s \end{gathered}$$

After 0.639 s drop will drop on the floor.

The time intervals between all the drops are same.

Between those 4 drops there are three equal intervals of time.

Therefore,

$$\begin{gathered} t=\frac{0.639}{3} \\ =0.213\mathrm{s} \end{gathered}$$

Time of second drop leaving the nozzle after the first drop is 0.213 s.

Time of third drop leaving the nozzle after the second drop is 0.213 s.

Time of second drop and nozzle when first drop falls on the floor is,

$$\begin{gathered} t_2=0.693-0.426 \\ =0.213s \end{gathered}$$

Time of third drop and nozzle when first drop falls on floor is,

$$\begin{gathered} t_3=0.693-0.426 \\ =0.213s \end{gathered}$$

To calculate the distance of the second drop from the nozzle, use the time taken by the second drop to fall on the ground. The time is 0.426 s . Therefore,

$$\begin{gathered} ∆\mathrm{y}={\mathrm{v}}_1\mathrm{t}+0.5{\mathrm{at}}^2 \\ {\mathrm{y}}^2=\left(0.5\right)\left(-9.8\right)\left(0.{426}^2\right) \\ =-0.889\mathrm{m} \end{gathered}$$

Therefore, the second drop is 0.889 m below the nozzle.

Similarly, to calculate the distance of the third drop from the nozzle, use the time taken by the third drop to fall on the ground. The time is 0.213 s . Therefore,

$$\begin{gathered} ∆y=v_{1}t+0.5a{t}^2 \\ {y}^3=\left(0.5\right)\left(-9.8\right)\left(0.{213}^2\right) \\ =-0.222m \end{gathered}$$

Therefore, third drop is 0.222 m below the nozzle.