Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 2: Problem 54

please assume the free-fall acceleration \(g=9.80 \mathrm{m} / \mathrm{s}^{2}\) unless a more precise value is given in the problem statement. Ignore air resistance. A student, looking toward his fourth-floor dormitory window, sees a flowerpot with nasturtiums (originally on a window sill above) pass his \(2.0-\mathrm{m}\) high window in 0.093 s. The distance between floors in the dormitory is $4.0 \mathrm{m} .$ From a window on which floor did the flowerpot fall?

Short Answer

Expert verified
Answer: The flowerpot was dropped from the fourth-floor window.

Step by step solution

01

Identify given data

Here are the given data: - Time taken to pass the window: \(t = 0.093 \ \text{s}\) - Height of the student's window: \(h_\text{window} = 2.0 \ \text{m}\) - Distance between floors: \(d_\text{floors} = 4.0 \ \text{m}\) - Free-fall acceleration: \(g = 9.80 \ \frac{\text{m}}{\text{s}^2}\)
02

Calculate the velocity when the flowerpot reaches the top of the window

We'll use the following kinematic equation to find the velocity \(v_\text{top}\) when the flowerpot reaches the top of the window: \(v^2 = u^2 + 2as\) Where: - \(v\) is the final velocity - \(u\) is the initial velocity (in this case, 0 m/s as it starts from rest) - \(a\) is the acceleration (in this case, -g, as the flowerpot is in free-fall) - \(s\) is the displacement (in this case, the height of the window) \(v_\text{top}^2 = 0^2 + 2(-g)(h_\text{window})\) \(v_\text{top} = \sqrt{2(-g)(h_\text{window})}\) \(v_\text{top} = \sqrt{2(-(-9.80))(2.0)}\) \(v_\text{top} = \sqrt{39.2}\) \(v_\text{top} \approx 6.26 \ \frac{\text{m}}{\text{s}}\)
03

Calculate the total height the flowerpot fell from

Now we'll use another kinematic equation to calculate the height \(h_\text{total}\), taking the calculated velocity at the top of the window: \(h_\text{total} = ut + \frac{1}{2}at^2\) Where: - \(h_\text{total}\) is the total height fallen - \(u\) is the initial velocity (in this case, 0 m/s as it starts from rest) - \(t\) is the time taken to pass the window - \(a\) is the acceleration (in this case, -g, as the flowerpot is in free-fall) \(h_\text{total} = 0 \cdot t + \frac{1}{2}(-g)t^2\) \(h_\text{total} = -\frac{1}{2}(9.80)(0.093)^2\) \(h_\text{total} \approx 0.0423 \ \text{m}\)
04

Determine the initial height where it started falling from

Now, adding the height of the window to the height calculated in step 3, we'll find the initial height: \(h_\text{initial} = h_\text{total} + h_\text{window}\) \(h_\text{initial} = 0.0423 + 2.0\) \(h_\text{initial} \approx 2.0423 \ \text{m}\)
05

Determine the floor the flowerpot fell from

Now, we'll divide the initial height by the distance between the floors: \(N_\text{floors} = \frac{h_\text{initial}}{d_\text{floors}}\) \(N_\text{floors} = \frac{2.0423}{4.0}\) \(N_\text{floors} \approx 0.5106\) Since the result is about half a floor above the student's floor (the 0.5106 value), the flowerpot must have fallen from one floor above the student, so it was dropped from his fourth-floor window.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Jason drives due west with a speed of \(35.0 \mathrm{mi} / \mathrm{h}\) for 30.0 min, then continues in the same direction with a speed of $60.0 \mathrm{mi} / \mathrm{h}\( for \)2.00 \mathrm{h},$ then drives farther west at \(25.0 \mathrm{mi} / \mathrm{h}\) for \(10.0 \mathrm{min} .\) What is Jason's average velocity for the entire trip?
In the problems, please assume the free-fall acceleration $g=9.80 \mathrm{m} / \mathrm{s}^{2}$ unless a more precise value is given in the problem statement. Ignore air resistance. A stone is thrown vertically downward from the roof of a building. It passes a window \(16.0 \mathrm{m}\) below the roof with a speed of $25.0 \mathrm{m} / \mathrm{s} .\( It lands on the ground \)3.00 \mathrm{s}$ after it was thrown. What was (a) the initial velocity of the stone and (b) how tall is the building?
A ball thrown by a pitcher on a women's softball team is timed at 65.0 mph. The distance from the pitching rubber to home plate is \(43.0 \mathrm{ft}\). In major league baseball the corresponding distance is \(60.5 \mathrm{ft}\). If the batter in the softball game and the batter in the baseball game are to have equal times to react to the pitch, with what speed must the baseball be thrown? Assume the ball travels with a constant velocity. [Hint: There is no need to convert units; set up a ratio.]
In the problems, please assume the free-fall acceleration $g=9.80 \mathrm{m} / \mathrm{s}^{2}$ unless a more precise value is given in the problem statement. Ignore air resistance. To pass a physical fitness test, Marcella must run \(1000 \mathrm{m}\) at an average speed of \(4.00 \mathrm{m} / \mathrm{s} .\) She runs the first $500 \mathrm{m}\( at an average of \)4.20 \mathrm{m} / \mathrm{s} .$ (a) How much time does she have to run the last \(500 \mathrm{m} ?\) (b) What should be her average speed over the last \(500 \mathrm{m}\) in order to finish with an overall average speed of \(4.00 \mathrm{m} / \mathrm{s} ?\)
A 1200 -kg airplane starts from rest and moves forward with a constant acceleration of magnitude \(5.00 \mathrm{m} / \mathrm{s}^{2}\) along a runway that is 250 m long. (a) How long does it take the plane to reach a speed of \(46.0 \mathrm{m} / \mathrm{s} ?\) (b) How far along the runway has the plane moved when it reaches \(46.0 \mathrm{m} / \mathrm{s} ?\)
See all solutions

Recommended explanations on Physics Textbooks

Geometrical and Physical Optics

Read Explanation

Classical Mechanics

Read Explanation

Famous Physicists

Read Explanation

Fields in Physics

Read Explanation

Turning Points in Physics

Read Explanation

Torque and Rotational Motion

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free