/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 84 A flowerpot falls off a windowsi... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A flowerpot falls off a windowsill and falls past the window below. You may ignore air resistance. It takes the pot 0.420 s to pass from the top to the bottom of this window, which is 1.90 \(\mathrm{m}\) high. How far is the top of the window below the windowsill from which the flowerpot fell?

Short Answer

Expert verified
The top of the window is approximately 0.31 meters below the windowsill.

Step by step solution

01

Understand the Problem

A flowerpot falls off a windowsill and passes a window that is 1.90 m tall in 0.420 s. We want to find how far the top of the window is below the windowsill.
02

Establish Known Variables

- Height of the window, \( h = 1.90 \; \text{m} \)- Time taken to pass the window, \( t = 0.420 \; \text{s} \)- Acceleration due to gravity, \( g = 9.81 \; \text{m/s}^2 \)We need to find the initial distance the pot falls before reaching the top of the window.
03

Use Kinematic Equation for Final Velocity

Since the final velocity \( v_f \) when the pot reaches the bottom of the window is unknown, we use the kinematic equation \( s = ut + \frac{1}{2}at^2 \) from the top to the bottom of the window where \( s = 1.90 \; \text{m} \), initial velocity \( u \) is not known, and acceleration \( a = g = 9.81 \; \text{m/s}^2 \). We get:\[ 1.90 = u \times 0.42 + \frac{1}{2} \times 9.81 \times (0.42)^2 \]
04

Solve for Initial Velocity at Top of Window

Rearrange the equation from Step 3:\[ 1.90 = u \times 0.42 + 0.5 \times 9.81 \times 0.1764 \]\[ 1.90 = u \times 0.42 + 0.864396 \]Solving for \( u \):\[ u = \frac{1.90 - 0.864396}{0.42} \approx 2.465 \; \text{m/s} \]
05

Calculate the Initial Drop Distance Using Velocity

Now, use the kinematic equation from the windowsill to the top of the window where initial velocity is 0, distance \( s \) is what we are trying to find, and final velocity is 2.465 m/s:Using \( v_f^2 = u^2 + 2as \):\[ (2.465)^2 = 0 + 2 \times 9.81 \times s \]\[ 6.076225 = 19.62s \]\[ s = \frac{6.076225}{19.62} \approx 0.31 \; \text{m} \]
06

Final Solution

The top of the window is approximately 0.31 meters below the windowsill from which the flowerpot fell.

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.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Motion under Gravity
When an object is in free fall, like a flowerpot falling from a window, it is moving under the influence of gravity alone. On Earth, this gravitational force accelerates objects downwards at approximately 9.81 m/s². This means every second, its velocity increases by 9.81 m/s.
This concept of motion under gravity is crucial in understanding how objects behave when they fall. We assume there is no air resistance, meaning that each falling object's motion solely results from gravity. This assumption simplifies calculations and lets us use specific kinematic equations to predict the object's behavior.
It's essential to know that motion under gravity affects all objects equally if no other forces are acting upon them, meaning they all accelerate similarly regardless of their masses.
Kinematic Equations
Kinematic equations describe the motion of objects. They are particularly helpful for understanding motion under constant acceleration, such as motion under gravity. These equations allow us to calculate unknown variables like distance, final velocity, and time, using known variables.
Three primary kinematic equations are most commonly used:
  • \( v = u + at \)
  • \( s = ut + \frac{1}{2}at^2 \)
  • \( v^2 = u^2 + 2as \)
- \( v \) is the final velocity.- \( u \) is the initial velocity.- \( a \) is the acceleration.- \( t \) is the time.- \( s \) is the displacement or distance.
In the flowerpot example, these equations help us find how far it fell before passing the window by recognizing that the pot accelerated uniformly downwards.
Initial Velocity Calculation
To solve problems with motion under gravity, calculating the initial velocity can be necessary, as it was in the flowerpot example. Even if we don't initially know this value, it can be derived using kinematic equations.
By rearranging the kinematic equation \( s = ut + \frac{1}{2}at^2 \), where \( s \) is the distance travelled, we can solve for the initial velocity \( u \). Given enough information about the other quantities involved, manipulating these equations allows us to isolate and find the unknown \( u \).
In our example, the initial velocity was determined through analyzing how fast the flowerpot was moving at the top of the window. Once we have this value, it gives us enough information to backtrack and compute the distance it fell from the windowsill to the window's top.

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

At the instant the traffic light turns green, a car that has been waiting at an intersection starts ahead with a constant acceleration of 3.20 \(\mathrm{m} / \mathrm{s}^{2} .\) At the same same instant a truck, traveling with a constant speed of \(20.0 \mathrm{m} / \mathrm{s},\) overtakes and passes the car. (a) How far beyond its starting point does the car overtake the truck? (b) How fast is the car traveling when it overtakes the truck? (c) Sketch an \(x-t\) graph of the motion of both vehicles. Take \(x=0\) at the intersection. (d) Sketch a \(v_{x}-t\) graph of the motion of both vehicles.

A bird is flying due east. Its distance from a tall building is given by \(x(t)=28.0 \mathrm{m}+(12.4 \mathrm{m} / \mathrm{s}) t-\left(0.0450 \mathrm{m} / \mathrm{s}^{3}\right) t^{3}\) What is the instantaneous velocity of the bird when \(t=8.00 \mathrm{s}?\)

A subway train starts from rest at a station and accelerates at a rate of 1.60 \(\mathrm{m} / \mathrm{s}^{2}\) for 14.0 \(\mathrm{s}\) . It runs at constant speed for 70.0 \(\mathrm{s}\) and slows down at a rate of 3.50 \(\mathrm{m} / \mathrm{s}^{2}\) until it stops at the next station. Find the total distance covered.

Determined to test the law of gravity for himself, a student walks off a skyscraper 180 \(\mathrm{m}\) high, stopwatch in hand, and starts his free fall (zero initial velocity). Five seconds later, Superman arrives at the scene and dives off the roof to save the student. Superman leaves the roof with an initial speed \(v_{0}\) that he produces by pushing himself downward from the edge of the roof with his legs of steel. He then falls with the same acceleration as any freely falling body. (a) What must the value of \(v_{0}\) be so that Superman catches the student just before they reach the ground? (b) On the same graph, sketch the positions of the student and of Superman as functions of time. Take Superman's initial speed to have the value calculated in part (a). (c) If the height of the skyscraper is less than some minimum value, even Superman can't reach the student before he hits the ground. What is this minimum height?

Launch of the Space Shuttle. At launch the space shuttle weighs 4.5 million pounds. When it is launched from rest, it takes 8.00 s to reach 161 \(\mathrm{km} / \mathrm{h}\) , and at the end of the first 1.00 \(\mathrm{min}\) its speed is 1610 \(\mathrm{km} / \mathrm{h}\) . (a) What is the average acceleration (\(\operatorname{in}\) \(\mathrm{m} / \mathrm{s}^{2}\) ) of the shuttle (i) during the first \(8.00 \mathrm{s},\) and (ii) between 8.00 \(\mathrm{s}\) and the end of the first 1.00 \(\mathrm{min}\) ? (b) Assuming the acceleration is constant during each time interval (but not necessarily the same in both intervals, what distance does the shuttle travel (i) during the first \(8.00 \mathrm{s},\) and ( ii) during the interval from 8.00 s to 1.00 \(\mathrm{min}\)?

See all solutions

Recommended explanations on Physics Textbooks

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.