Question

A dripping water faucet steadily releases drops 1.0 s apart. As these drops fall, does the distance between them increase, decrease, or remain the same? Prove your answer.

Answer 1

Answer:

Distance between them increase

Explanation:

The position S of the water droplet can be determined  using equation of motion

$S=ut+\frac{1}{2} at^2$

where $u$ is the initial velocity which is zero here

$t$ is time taken, $a$ is acceleration due to gravity

the position of  first drop after time $t$ is given by

$S_{1} =0 \times t+ \frac{1}{2} at^2=\frac{1}{2} at^2............(1)$

the position of  next drop at same time is

$S_{2} =\frac{1}{2} a(t-1)^2 = \frac{1}{2} a(t^2+1-2t)............(2)$

distance between them is $S_{1} -S_{2}$  is $a(t-1)$

from the above the difference will increase with the time