/*! 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 74 A flowerpot falls off a windowsi... [FREE SOLUTION] | 91Ó°ÊÓ

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A flowerpot falls off a windowsill and passes the window of the story below. Ignore air resistance. It takes the pot \(0.380 \mathrm{~s}\) 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 starting height or the distance from the windowsill to the top of the window can be found by summing the height of the fall before the window and the height of the window itself.

Step by step solution

01

Identify the variables

From the problem, we can identify the following variables for the distance between opening and closing of the window: \(h = 1.90 \mathrm{~m}\) (height of the window), \(t = 0.380 \mathrm{~s}\) (time to cross the window), \(g = 9.8 \mathrm{~m/s}^2\) (acceleration due to gravity).
02

Calculate Initial velocity

We don't know the initial velocity of the pot when it reached at the top of window. We can calculate this using the kinematic equation of motion, which states: \(2gs = v^2 - u^2\). Since the flowerpot is moving downwards, \(s = -h\), \(u = 0 \mathrm{~m/s}\) and \(v\), the final velocity is what we’re solving for. Rearranging and solving gives us \(v = \sqrt{2gh}\).
03

Calculate height

Using another kinematic equation of motion \(s = ut + \frac{1}{2}gt^2\), we know that the pot had some initial velocity when it reached top of the window that we calculated in step 2. Plugging these values back into the equation will allow us to calculate the falling distance before it reaches at the top of the window, this will be \(s = v*t+\frac{1}{2}g*t^2\) where variable \(s\) will be the distance from windowsill to the top of window.
04

Calculate total fall

Finally, we total the height of the window and the fall before the window. This will give us the total distance the pot fell from the windowsill. This will be : \(h_{total} = h + s\).

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Key Concepts

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

Free Fall Physics
Free fall physics is a fundamental concept that deals with the motion of objects under the sole influence of gravity. This scenario assumes negligible air resistance, meaning the only force acting on the falling object is due to gravity.

When an object like a flowerpot is dropped from a height, it starts from rest, and its speed increases as it descends, because it is being accelerated by Earth's gravitational pull. The speed at which the object falls, and the distance it covers, are determined by kinematic equations. An important aspect of free fall is that all objects, regardless of their mass, will have the same acceleration due to gravity, typically represented by the symbol \( g \), with a standard value of approximately \( 9.8 \text{m/s}^2 \) on the surface of Earth.

In the absence of air resistance, the problem simplifies significantly. We can accurately predict the object's motion using the provided kinematic equations. Understanding free fall is vital, not just for solving physics problems, but also in real-life applications such as calculating the impact force of falling objects or the trajectories of projectiles.
Initial Velocity Calculation
The initial velocity of an object is the speed at which it begins its motion. In the context of free fall, it is common to consider the object as starting from rest, which translates to an initial velocity \( u = 0 \text{m/s} \). However, when considering an object like the flowerpot already in motion, calculating its initial velocity as it passes a given point (like the top of the window) requires the use of kinematic equations.

These equations relate initial velocity, final velocity, acceleration, distance, and time. For our flowerpot, we calculate the velocity at the top of the window using the equation \( v = \sqrt{2gh} \), where \( g \) is the acceleration due to gravity and \( h \) is the height of the window. Despite the flowerpot being in motion before reaching the window, for the brief period of passing the window, we consider the top of the window as the starting point hence a new 'initial velocity' for our calculations which is the velocity at the very top of the window.
Acceleration Due to Gravity
Acceleration due to gravity is a measure of the rate at which an object increases its velocity due to the force of Earth's gravity. On the surface of the Earth, this value is approximately \( 9.8 \text{m/s}^2 \), and it is considered to be constant over short distances such as the height of a building. This acceleration is what causes freely falling objects to increase their speed as they descend.

In kinematic calculations, the uniform acceleration provided by gravity simplifies problem-solving. The acceleration does not depend on the mass of the object, meaning that a feather and a hammer would fall at the same rate in a vacuum. Since our flowerpot is in free fall, it accelerates at this constant rate from the moment it leaves the windowsill until it passes the window below. The term \( g \) appears in several kinematic equations that describe the motion of the flowerpot, and it's crucial for predicting how the pot falls and the velocity it gains over time.

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Most popular questions from this chapter

A student is running at her top speed of \(5.0 \mathrm{~m} / \mathrm{s}\) to catch a bus, which is stopped at the bus stop. When the student is still \(40.0 \mathrm{~m}\) from the bus, it starts to pull away, moving with a constant acceleration of \(0.170 \mathrm{~m} / \mathrm{s}^{2}\). (a) For how much time and what distance does the student have to run at \(5.0 \mathrm{~m} / \mathrm{s}\) before she overtakes the bus? (b) When she reaches the bus, how fast is the bus traveling? (c) Sketch an \(x-t\) graph for both the student and the bus. Take \(x=0\) at the initial position of the student. (d) The equations you used in part (a) to find the time have a second solution, corresponding to a later time for which the student and bus are again at the same place if they continue their specified motions. Explain the significance of this second solution. How fast is the bus traveling at this point? (e) If the student's top speed is \(3.5 \mathrm{~m} / \mathrm{s},\) will she catch the bus? (f) What is the minimum speed the student must have to just catch up with the bus? For what time and what distance does she have to run in that case?

(a) If a flea can jump straight up to a height of \(0.440 \mathrm{~m}\), what is its initial speed as it leaves the ground? (b) How long is it in the air?

During an auto accident, the vehicle's airbags deploy and slow down the passengers more gently than if they had hit the windshield or steering wheel. According to safety standards, airbags produce a maximum acceleration of \(60 g\) that lasts for only \(36 \mathrm{~ms}\) (or less). How far (in meters) does a person travel in coming to a complete stop in \(36 \mathrm{~ms}\) at a constant acceleration of \(60 \mathrm{~g}\) ?

A tennis ball on Mars, where the acceleration due to gravity is \(0.379 g\) and air resistance is negligible, is hit directly upward and returns to the same level \(8.5 \mathrm{~s}\) later. (a) How high above its original point did the ball go? (b) How fast was it moving just after it was hit? (c) Sketch graphs of the ball's vertical position, vertical velocity, and vertical acceleration as functions of time while it's in the Martian air.

A lunar lander is making its descent to Moon Base I (Fig. E2.40). The lander descends slowly under the retro-thrust of its descent engine. The engine is cut off when the lander is \(5.0 \mathrm{~m}\) above the surface and has a downward speed of \(0.8 \mathrm{~m} / \mathrm{s}\). With the engine off, the lander is in free fall. What is the speed of the lander just before it touches the surface? The acceleration due to gravity on the moon is \(1.6 \mathrm{~m} / \mathrm{s}^{2}\).

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