/*! 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 45 From her bedroom window a girl d... [FREE SOLUTION] | 91Ó°ÊÓ

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From her bedroom window a girl drops a water-filled balloon to the ground, \(6.0 \mathrm{~m}\) below. If the balloon is released from rest, how long is it in the air?

Short Answer

Expert verified
The balloon is in the air for approximately 1.11 seconds.

Step by step solution

01

Write Down Known Variables

First, identify the variables from the problem. We need to calculate the time the balloon is in the air. The distance from the window to the ground (height) is given as \(h = 6.0 \text{ m}\). The initial velocity \(u = 0 \text{ m/s}\) since it is released from rest. The acceleration due to gravity \(g = 9.8 \text{ m/s}^2\).
02

Use the Kinematic Equation for Free Fall

The kinematic equation that relates distance with time for an object in free fall is:\[h = ut + \frac{1}{2}gt^2\]where:- \(h\) is the height (6.0 m),- \(u\) is the initial velocity (0 m/s),- \(g\) is the acceleration due to gravity (9.8 m/s²),- \(t\) is the time in seconds.
03

Substitute Known Values into the Equation

Substitute the known values into the equation:\[6.0 = 0 \cdot t + \frac{1}{2} \cdot 9.8 \cdot t^2\]This simplifies to:\[6.0 = 4.9t^2\]
04

Solve for Time \(t\)

Isolate \(t^2\) by dividing both sides by 4.9:\[t^2 = \frac{6.0}{4.9}\]Calculate the division:\[t^2 \approx 1.224\]Take the square root of both sides to solve for \(t\):\[t \approx \sqrt{1.224} \approx 1.106\]
05

Round Your Answer

Since the question typically asks for time to a reasonable number of significant figures or decimal places, let's round \(t\) to two decimal places:\[t \approx 1.11 \text{ s}\]

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Key Concepts

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

Kinematic Equations
When dealing with motion in physics, kinematic equations are invaluable tools to help us predict the future behavior of moving objects. They describe the motion of objects under constant acceleration without needing to consider the forces causing the motion. For objects in free fall, which are only influenced by gravity, these equations become particularly handy. In this scenario, we often use the formula:\[h = ut + \frac{1}{2}gt^2\]Here, each symbol represents a specific component of the motion:
  • \(h\) is the height the object falls or travels.
  • \(u\) is the initial velocity or speed at which the object starts.
  • \(g\) is the acceleration due to gravity.
  • \(t\) is the time elapsed.
By manipulating these equations, we can solve for various components like time, height, or velocity, provided we have enough information. Free-falling objects, like the water balloon from the scenario, are excellent examples of real-life applications of these kinematic equations.
Acceleration due to Gravity
Gravity is a universal force that attracts objects toward each other. On Earth, this force gives objects an acceleration of approximately \(9.8 \text{ m/s}^2\). This value is known as the "acceleration due to gravity" and is a crucial factor in solving free fall problems.
Objects in free fall are impacted by this constant acceleration, causing their speed to increase as they drop.
Because of gravity:
  • The velocity of a freely falling object increases by about \(9.8 \text{ m/s}\) for every second it is in the air.
  • An object in free fall will cover more distance with each passing second, as its speed builds up.
Understanding this acceleration simplifies solving for time, distance, and velocity in free fall scenarios. As seen in the exercise, knowing the acceleration due to gravity allowed us to use the kinomatic equation directly to find the time the balloon was in the air.
Initial Velocity
Initial velocity is the speed at which an object begins its motion. In many free-fall problems like the one given, the object starts from rest, meaning its initial velocity is zero.
When the initial velocity \(u\) is zero, our kinematic equations simplify, often leading to easier calculations.
  • There is no initial movement to consider, so the initial velocity term \(u \cdot t\) in the equation \(h = ut + \frac{1}{2}gt^2\) drops out.
  • This simplification means that the formula directly calculates the motion influenced only by gravity.
Thus, for a balloon dropped from rest, understanding that the initial velocity is zero makes it straightforward to apply the equation to find the time it takes to fall a specific height.

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Most popular questions from this chapter

A ball is thrown upward from the top of a 25.0 -m-tall building. The ball's initial speed is \(12.0 \mathrm{~m} / \mathrm{s}\). At the same instant, a person is running on the ground at a distance of \(31.0 \mathrm{~m}\) from the building. What must be the average speed of the person if he is to catch the ball at the bottom of the building?

A runner accelerates to a velocity of \(4.15 \mathrm{~m} / \mathrm{s}\) due west in \(1.50 \mathrm{~s} .\) His average acceleration is \(0.640 \mathrm{~m} / \mathrm{s}^{2},\) also directed due west. What was his velocity when he began accelerating?

A speedboat starts from rest and accelerates at \(+2.01 \mathrm{~m} / \mathrm{s}^{2}\) for \(7.00 \mathrm{~s}\). At the end of this time, the boat continues for an additional \(6.00 \mathrm{~s}\) with an acceleration of \(+0.518 \mathrm{~m} / \mathrm{s}^{2}\) Following this, the boat accelerates at \(-1.49 \mathrm{~m} / \mathrm{s}^{2}\) for \(8.00 \mathrm{~s}\). (a) What is the velocity of the boat at \(t=21.0 \mathrm{~s} ?\) (b) Find the total displacement of the boat.

Two runners start one hundred meters apart and run toward each other. Each runs ten meters during the first second. During each second thereafter, each runner runs ninety percent of the distance he ran in the previous second. Thus, the velocity of each person changes from second to second. However, during any one second, the velocity remains constant. Make a position-time graph for one of the runners. From this graph, determine (a) how much time passes before the runners collide and (b) the speed with which each is running at the moment of collision.

Two stones are thrown simultaneously, one straight upward from the base of a cliff and the other straight downward from the top of the cliff. The stones are thrown with the same speed. (a) Does the stone thrown upward gain or lose speed as it moves upward? Why? (b) Does the stone thrown downward gain or lose speed as time passes? Explain. (c) The speed at which the stones are thrown is such that they cross paths. Where do they cross paths, above, at, or below the point that corresponds to half the height of the cliff? Justify your answer. Problem The height of the cliff is \(6.00 \mathrm{~m},\) and the speed with which the stones are thrown is \(9.00 \mathrm{~m} / \mathrm{s}\). Find the location of the crossing point. Check to see that your answer is consistent with your answers to the Concept Questions.

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