/*! 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 41 A wrecking ball is hanging at re... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A wrecking ball is hanging at rest from a crane when suddenly the cable breaks. The time it takes for the ball to fall halfway to the ground is \(1.2 \mathrm{~s}\). Find the time it takes for the ball to fall from rest all the way to the ground.

Short Answer

Expert verified
The wrecking ball takes approximately 1.7 seconds to fall to the ground.

Step by step solution

01

Analyze the Initial Situation

A wrecking ball is dropped from rest, which means its initial velocity, \( v_0 = 0 \). The ball falls under the influence of gravity, described by the acceleration \( g = 9.8 \, \text{m/s}^2 \). Given that the time to fall halfway is \( t_{1/2} = 1.2 \, \text{s} \).
02

Calculate Halfway Distance

Using the equation for distance under uniform acceleration, \( s = v_0t + \frac{1}{2}gt^2 \), compute the distance to the halfway point \( s_{1/2} \) by plugging in \( t = 1.2 \, \text{s} \):\[ s_{1/2} = 0 + \frac{1}{2} \cdot 9.8 \, \text{m/s}^2 \cdot (1.2 \, \text{s})^2 = 7.056 \, \text{m} \]
03

Set Up Full Distance Equation

Let the total distance the ball falls be \( D \). From the previous step, we have \( D/2 = 7.056 \, \text{m} \), so \( D = 14.112 \, \text{m} \). Use the distance equation for the full distance \( D \) to find the total fall time \( t \):\[ D = \frac{1}{2} \cdot g \cdot t^2 \]
04

Solve for Total Time

Re-arrange the distance equation to solve for \( t \):\[ t = \sqrt{\frac{2D}{g}} = \sqrt{\frac{2 \cdot 14.112}{9.8}} \approx 1.7 \, \text{s} \]

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.

Free fall motion
Free fall motion occurs when an object is falling under the sole influence of gravity, with no air resistance acting upon it. This special type of motion is characterized by a constant acceleration towards the Earth's surface, which makes it a fascinating topic in physics. In the context of our exercise, the wrecking ball falls freely once the cable snaps.
Imagine dropping an object off a tall building— as soon as it starts falling, it is in free fall. It doesn’t matter if it’s a feather or a wrecking ball (assuming no air resistance), the motion and characteristics are the same, driven purely by gravity. This means the velocity increases as it falls, but the rate of increase remains constant unless outside forces such as wind or air resistance interfere.
Understanding free fall helps us grasp how gravity influences motion across not just Earth, but any celestial body.
Acceleration due to gravity
Acceleration due to gravity is a fundamental concept in physics affecting all objects in free fall. On Earth, this acceleration has a nearly constant value of approximately 9.8 m/s², which means that for every second an object is in free fall, its velocity increases by 9.8 m/s.
  • It is this constant acceleration that defines the path of free-falling objects and allows us to predict how far or how fast an object is going after falling for a certain period of time.
  • This constant, often denoted by the symbol "g," makes calculations straightforward, allowing us to use equations of motion effectively.

For our wrecking ball problem, "g" plays a crucial role in determining how quickly the ball reaches the ground. By using this constant, it was calculated how far the ball falls in a given time, providing vital information that enabled the discovery of the total time taken for the complete fall.
Equations of motion
Equations of motion are mathematical tools that describe how objects move under the influence of various forces, such as gravity. They are particularly useful in kinematics for determining unknown variables when an object is in motion. The primary set of equations that describe motion with constant acceleration (a key feature of free fall) include:
  • The first equation relates velocity, acceleration, and time: \( v = v_0 + at \)
  • The second equation involves displacement, initial velocity, time, and acceleration: \( s = v_0t + \frac{1}{2}at^2 \)
  • The third links velocity, initial velocity, acceleration, and displacement: \( v^2 = v_0^2 + 2as \)

In solving our problem, the second equation was crucial, allowing us to find how far the ball fell in a given time and, subsequently, how long the total fall to the ground took. By plugging in the known values, the straightforward nature of these equations becomes apparent, making them invaluable for solving real-world physics problems.

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

A whale swims due east for a distance of \(6.9 \mathrm{~km},\) turns around and goes due west for \(1.8 \mathrm{~km},\) and finally turns around again and heads \(3.7 \mathrm{~km}\) due east. (a) What is the total distance traveled by the whale? (b) What are the magnitude and direction of the displacement of the whale?

Interactive Solution \(\underline{2.31}\) at offers help in modeling this problem. A car is traveling at a constant speed of \(33 \mathrm{~m} / \mathrm{s}\) on a highway. At the instant this car passes an entrance ramp, a second car enters the highway from the ramp. The second car starts from rest and has a constant acceleration. What acceleration must it maintain, so that the two cars meet for the first time at the next exit, which is \(2.5 \mathrm{~km}\) away?

A VW Beetle goes from 0 to \(60.0 \mathrm{mi} / \mathrm{h}\) with an acceleration of \(+2.35 \mathrm{~m} / \mathrm{s}^{2}\). (a) How much time does it take for the Beetle to reach this speed? (b) A top-fuel dragster can go from 0 to \(60.0 \mathrm{mi} / \mathrm{h}\) in \(0.600 \mathrm{~s}\). Find the acceleration \(\left(\mathrm{in} \mathrm{m} / \mathrm{s}^{2}\right)\) of the dragster.

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.

A ball is dropped from rest from the top of a cliff that is \(24 \mathrm{~m}\) high. From ground level, a second ball is thrown straight upward at the same instant that the first ball is dropped. The initial speed of the second ball is exactly the same as that with which the first ball eventually hits the ground. In the absence of air resistance, the motions of the balls are just the reverse of each other. Determine how far below the top of the cliff the balls cross paths.

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.