Question

Water drips from the nozzle of a shower onto the floor 200 cm 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 (b) third drops?

Answer 1

Answer:

a) 89cm

b) 23cm

Explanation:

How much time take for the drop to strick the floor?

From Kinematics the position as a function of time is:

$y(t)=y_i-\frac{1}{2}gt^2$

where

$y_i=200cm=2m$

$g=9.8\frac{m}{s^2}$ is gravity's acceleration

We want the time for which y(t)=0, so

$y_i-\frac{1}{2}gt^2=0$ ⇒ $\frac{1}{2}gt^2=y_i$

⇒ $t=\sqrt{\frac{2y_i}{g} }$ ⇒ $t_1=0.639s$

This is the time it takes for the first drop to touch the floor.

Since the drops fall at regular times, and we know that at a time $t_1$ there are two drops in the air, the time between one drop and the next one to pass any point will be $\frac{t_1}{3}$.

So, the time since the second drop fallen is:

$t_2=2\frac{t_1}{3} =0.426s$

and the time since the third drop fallen is:

$t_3=\frac{t_1}{3} =0.213s$

So, now, the position for each drop is:

$y_2(t_2)=y_i-\frac{1}{2}gt_2^2=1.11m\\y_3(t_3)=y_i-\frac{1}{2}gt_3^2=1.77m$.

The question is how far below the nozzle. So:

$y_2^*=2m-1.11m=0.89m=89cm\\y_3^*=2m-1.77m=0.23m=23cm$