Question

Water drips from the nozzle of a shower onto the floor 193 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: 108.81 cm and 48.66 cm

Step-by-step explanation:

In this, we have to make sure to keep in mind the Gravity effects on the drops. The drops will accelerate when they fall making them travel faster. This means, the velocity is not constant.

What is know:

Height (h) = 193 cm

Gravity (g) = 981
`cm/s^(2)`

Initial Velocity = 0

First, we can know how long it take to the drop to travel to the floor. It can be done with the following equation:

`x = V_(0) t + (1)/(2) at^(2)` (1)

Where:

x is the distance which is 193cm

Vo is the Initial Velocity which is zero

t is the time the time it takes the drop to travel from the shower to the floor

a is the aceleration, which in this case is the gravity.

With the Initial Velocity equals zero the equations simply:

`193 cm = (1)/(2)gt^(2)`

To search for the time:

`t =\sqrt{(2*193cm)/(981cm/s^(2) ) }`

t = 0.627 s

This is the time it takes a drop to fall to the floor, with this time and knowing other 3 drops have driped from the shower by this time. We can calculate how much time it takes the shower to drip each drop.

Time for Drip = t/4

Time for Drip = 0.156

This time is the difference between each drop, using the same equation we can calculate where was each drop, because now it is know how much time had each drop after being drip from the shower.

Our first is already on the floor (193 cm) with 0.627 s, The second drops have been falling for (0.627s - 0.156) 0.471 s and our third drop for (0.627s - 0.156 - 0.156) 0.315 s

We can use (1) to know how far have each drop traveled on these times. We know the Initials Velocity are 0, know we need ot know the distances.

For the second drop:

`x = (1)/(2) (981cm/s^(2))(0.471s)^(2)`

`x =108.81 cm`

For the third drop:

`x = (1)/(2) (981cm/s^(2))(0.315s)^(2)`

`x = 48.66 cm`