/*! 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 53 A key falls from a bridge that i... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 53

A key falls from a bridge that is \(45 \mathrm{~m}\) above the water. It falls directly into a model boat, moving with constant velocity, that is \(12 \mathrm{~m}\) from the point of impact when the key is released. What is the speed of the boat?

Short Answer

Expert verified
The boat's speed is approximately 3.96 m/s.

Step by step solution

01

Calculate the Time for the Key to Fall

We start by calculating the time it takes for the key to fall 45 meters. Using the formula for the time of free fall, \[ t = \sqrt{\frac{2h}{g}} \]where \( h \) is the height (45 m) and \( g \) is the acceleration due to gravity (approx. 9.8 m/s虏). Substitute the values:\[ t = \sqrt{\frac{2 \times 45}{9.8}} \]Calculating further, \[ t = \sqrt{\frac{90}{9.8}} \approx \sqrt{9.1837} \approx 3.03 \text{ seconds} \]Thus, it takes approximately 3.03 seconds for the key to hit the water.
02

Determine the Boat's Constant Speed

Since the boat moves at constant speed and has 12 meters to travel in the time it takes the key to fall (3.03 seconds), we can find the speed using the formula:\[ v = \frac{d}{t} \]where \( d \) is the distance (12 meters) and \( t \) is the time (3.03 seconds). Substituting the values:\[ v = \frac{12}{3.03} \approx 3.96 \text{ m/s} \]Thus, the speed of the boat is approximately 3.96 m/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 91影视!

Key Concepts

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

Free Fall
Free fall refers to the motion of an object solely under the influence of gravity. In this case, only gravity acts on the key, with no air resistance considered. When an object is in free fall, its initial velocity is zero (if it starts from rest), and it accelerates downwards at a constant rate due to gravity.

In the exercise, the key falls from a 45-meter-high bridge into the water below. The formula used to determine the time of free fall is: \[ t = \sqrt{\frac{2h}{g}} \]where:
  • \( h \) is the height from which the key falls, (45 m in this case).
  • \( g \) is the acceleration due to gravity, approximated as \( 9.8 \, \text{m/s}^2 \).
This formula stems from the kinematic equation for uniformly accelerated motion, simplifying the process by assuming initial velocity is zero. After plugging in the values, the calculated fall time is approximately \( 3.03 \) seconds.
Constant Velocity
Constant velocity means the speed of an object doesn't change as time passes. It travels equal distances in equal intervals of time. In the problem, the model boat moves at a constant speed throughout the key's fall.

To find the velocity, the distance the boat must travel is known, which is 12 meters, and the time it takes for the key to hit the water is 3.03 seconds. Using the formula: \[ v = \frac{d}{t} \]where:
  • \( v \) is the velocity of the boat.
  • \( d \) is the distance (12 m).
  • \( t \) is the time (3.03 s).
Plugging in these values, the constant velocity of the boat is approximately \( 3.96 \, \text{m/s} \), meaning the boat travels 3.96 meters every second under these conditions.
Acceleration due to Gravity
Acceleration due to gravity is an essential concept in physics, denoted by \( g \). It describes how gravity affects the motion of free-falling objects, pulling them towards Earth at a roughly constant rate of \( 9.8 \, \text{m/s}^2 \).

This value can slightly vary depending on where you are on the Earth, but it鈥檚 generally accepted as 9.8 m/s虏 for most calculations. In our scenario, this constant acceleration allows for the calculations involving the key's fall. The consistency of \( g \) simplifies predicting the behavior of falling objects, such as their fall time and impact speed.
  • It affects all objects equally in the absence of other forces, such as air resistance.
  • For free fall, it鈥檚 the only force considered acting on the object.
Thus, knowing \( g \) enables precise predictions about objects under the influence of gravity, as seen in the calculation of the fall time of the key.

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 hoodlum throws a stone vertically downward with an initial speed of \(15.0 \mathrm{~m} / \mathrm{s}\) from the roof of a building, \(30.0 \mathrm{~m}\) above the ground. (a) How long does it take the stone to reach the ground?

You are arguing over a cell phone while trailing an unmarked police car by \(25 \mathrm{~m}\); both your car and the police car are traveling at \(120 \mathrm{~km} / \mathrm{hr}\). Your argument diverts your attention from the police car for \(2.0 \mathrm{~s}\) (long enough for you to look at the phone and yell, "I won't do that!"). At the beginning of that \(2.0 \mathrm{~s}\), the police officer begins braking suddenly at \(5.0 \mathrm{~m} / \mathrm{s}^{2}\). (a) What is the separation between the two cars when your attention finally returns? Suppose that you take another \(0.40\) s to realize your danger and begin braking. (b) If you too brake at \(5.0 \mathrm{~m} / \mathrm{s}^{2}\), what is your speed when you hit the police car?

An electron moving along the \(x\) axis has a position given by \(x=16 t e^{-t} \mathrm{~m}\), where \(t\) is in seconds. How far is the electron from the origin when it momentarily stops?

To set a speed record in a measured (straight-line) distance \(d\), a race car must be driven first in one direction (in time \(t_{1}\) ) and then in the opposite direction (in time \(t_{2}\) ). (a) To eliminate the effects of the wind and obtain the car's speed \(v_{c}\) in a windless situation, should we find the average of \(d / t_{1}\) and \(d / t_{2}\) (method 1 ) or should we divide \(d\) by the average of \(t_{1}\) and \(t_{2}\) ? (b) What is the fractional difference in the two methods when a steady wind blows along the car's route and the ratio of the wind speed \(v_{w}\) to the car's speed \(v_{c}\) is \(0.0240\) ?

A stone is dropped into a river from a bridge \(53.6 \mathrm{~m}\) above the water. Another stone is thrown vertically down \(1.00 \mathrm{~s}\) after the first is dropped. The stones strike the water at the same time. (a) What is the initial speed of the second stone? (b) Plot velocity versus time on a graph for each stone, taking zero time as the instant the first stone is released.
See all solutions

Recommended explanations on Physics Textbooks

Solid State Physics

Read Explanation

Physical Quantities And Units

Read Explanation

Wave Optics

Read Explanation

Electricity

Read Explanation

Electrostatics

Read Explanation

Circular Motion and Gravitation

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.