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Chapter 2: Problem 59

Water drips from the nozzle of a shower onto the floor 200 \(\mathrm{cm}\) 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?

Short Answer

Expert verified
The second drop is 89.4 cm below, and the third drop is 22.35 cm below the nozzle.

Step by step solution

01

Understand the sequence of drop falls

The first drop hits the floor at the same time the fourth drop starts falling. This means the time taken for each drop to fall must be equal, and let's call this time interval \(t\). Hence, if the first drop falls for a time \(3t\) (as the fourth drop only starts falling when it hits the floor), it reaches the ground in this time.
02

Calculate the time taken by the first drop

The first drop travels 200 cm in time \(3t\). We can use the formula for distance covered under gravity, \(s = \frac{1}{2}gt^2\). Thus, 200 cm \(= \frac{1}{2}g(3t)^2\). With \(g \approx 980 \text{ cm/s}^2\), solve for \(3t\) to get \(3t = \sqrt{\frac{2 \cdot 200}{980}} \approx 0.64\) s.
03

Determine the time intervals for the second and third drops

The second drop falls for time \(2t\) when the first drop just hits the floor. This means it has been falling for \(\frac{2}{3} \times 0.64\) seconds. Similarly, the third drop falls for \(t = \frac{1}{3} \times 0.64\) seconds.
04

Calculate distances for the second and third drops

Use the equation \(s = \frac{1}{2}gt^2\) for distance. - For the second drop, \(s_2 = \frac{1}{2} \times 980 \times (2t)^2\). Use \(2t = 2/3 \times 0.64\). Therefore, \(s_2 = \frac{1}{2} \times 980 \times (0.4267)^2 \approx 89.4\) cm.- Similarly, for the third drop, \(s_3 = \frac{1}{2} \times 980 \times (1/3 \times 0.64)^2 \approx 22.35\) cm.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equations of Motion
Understanding the equations of motion is crucial for solving projectile motion problems. These equations describe how objects move under the influence of forces, such as gravity.

For linear motion with constant acceleration, we use a set of equations. The key one we often use in problems like this is:
\[ s = ut + \frac{1}{2}at^2 \]where:
  • \(s\) = displacement or distance traveled
  • \(u\) = initial velocity (for a drop starting from rest, \(u = 0\))
  • \(a\) = acceleration (in this case, gravity)
  • \(t\) = time
In the problem, water drops from a shower and falls due to gravity. Since they start from rest, the initial velocity, \(u\), is zero. The equation simplifies to:
\[ s = \frac{1}{2}gt^2 \]Here, \(g\) represents the acceleration due to gravity, and this equation helps us solve how far each drop has traveled over a given time.
Gravitational Acceleration
Gravitational acceleration is the rate at which an object speeds up as it falls freely toward the Earth. On Earth, this constant is approximately \(g = 9.8 \ \text{m/s}^2\). In our problem, this value is translated to \(980 \ \text{cm/s}^2\) because the distances are measured in centimeters.

Understanding gravitational acceleration is crucial for calculating how fast an object falls. Gravity acts uniformly on all objects, regardless of their weight or mass, meaning any two objects dropped from the same height will hit the ground at the same time if air resistance is negligible.

In the context of the exercise, gravitational acceleration helps us compute the time it takes for each water drop to travel from the nozzle to the floor. The equation used involves substituting this constant to determine distances fallen over specific periods.
Kinematic Equations
Kinematic equations involve the mathematical descriptions of motion without considering the forces causing it. For constant acceleration, these equations are foundational to analyze motion in physics.

When dealing with problems like the dripping water droplets, the kinematic equation used is \[ s = \frac{1}{2}gt^2 \]The simplicity of this equation lies in its application to vertical free-fall problems and makes it a powerful tool. For each droplet, knowing one piece of information (time or distance) lets us find the other due to the factors being so clearly defined.

The exercise showcases how each drop's fall time and distance are connected. By breaking down the motion into straightforward calculable steps, kinematic equations help us easily find where the second and third drops are as the first reaches the floor.

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Most popular questions from this chapter

Suppose a rocket ship in deep space moves with constant acceleration equal to \(9.8 \mathrm{~m} / \mathrm{s}^{2}\), which gives the illusion of normal gravity during the flight. (a) If it starts from rest, how long will it take to acquire a speed one-tenth that of light, which travels at \(3.0 \times 10^{8} \mathrm{~m} / \mathrm{s} ?(\mathrm{~b})\) How far will it travel in so doing?

A certain elevator cab has a total run of \(190 \mathrm{~m}\) and a maximum speed of \(305 \mathrm{~m} / \mathrm{min}\), and it accelerates from rest and then back to rest at \(1.22 \mathrm{~m} / \mathrm{s}^{2} .\) (a) How far does the cab move while accelerating to full speed from rest? (b) How long does it take to make the nonstop \(190 \mathrm{~m}\) run, starting and ending at rest?

A muon (an elementary particle) enters a region with a speed of \(5.00 \times 10^{6} \mathrm{~m} / \mathrm{s}\) and then is slowed at the rate of \(1.25 \times 10^{14} \mathrm{~m} / \mathrm{s}^{2}\). (a) How far does the muon take to stop? (b) Graph \(x\) versus \(t\) and \(v\) versus \(t\) for the muon.

A rock is dropped from a 100 -m-high cliff. How long does it take to fall (a) the first \(50 \mathrm{~m}\) and \((\mathrm{b})\) the second \(50 \mathrm{~m}\) ?

A car can be braked to a stop from the autobahn-like speed of \(200 \mathrm{~km} / \mathrm{h}\) in \(170 \mathrm{~m}\). Assuming the acceleration is constant, find its magnitude in (a) SI units and (b) in terms of \(g\). (c) How much time \(T_{b}\) is required for the braking? Your reaction time \(T_{r}\) is the time you require to perceive an emergency, move your foot to the brake, and begin the braking. If \(T_{r}=400 \mathrm{~ms}\), then \((\mathrm{d})\) what is \(T_{b}\) in terms of \(T_{r}\), and (e) is most of the full time required to stop spent in reacting or braking? Dark sunglasses delay the visual signals sent from the eyes to the visual cortex in the brain, increasing \(T_{r} .(\mathrm{f})\) In the extreme case in which \(T_{r}\) is increased by \(100 \mathrm{~ms}\), how much farther does the car travel during your reaction time?
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