/*! 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 9 A physics book slides off a hori... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 9

A physics book slides off a horizontal tabletop with a speed of 1.10 \(\mathrm{m} / \mathrm{s}\) . It strikes the floor in 0.350 \(\mathrm{s}\) . Ignore air resistance. Find (a) the height of the tabletop above the floor; (b) the horizontal distance from the edge of the table to the point where the book strikes the floor, (c) the horizontal and vertical components of the book's velocity, and the magnitude and direction of its velocity, just before the book reaches the floor. (d) Draw \(x-t, y-t, v_{x}-t,\) and \(v_{y}-t\) graphs for the motion.

Short Answer

Expert verified
The table height is 0.601 m, and the book lands 0.385 m horizontally away. Its final velocity is 3.61 m/s at 72.3掳 below horizontal.

Step by step solution

01

Analyze the Problem

The problem involves a book sliding off a table. We need to consider the projectile motion since the book experiences free fall after leaving the table edge. Important given data includes an initial horizontal velocity and the time of flight.
02

Calculate the Height of the Table

To find the height (\(h\)) from which the book falls, we use the formula for free fall: \[ h = \frac{1}{2} g t^2 \]Given: acceleration due to gravity \(g = 9.81 \ \mathrm{m/s^2}\), and time \( t = 0.350 \ \mathrm{s}\).\[ h = \frac{1}{2} \times 9.81 \times (0.350)^2 = 0.601 \ \mathrm{m} \] Thus, the height of the table is 0.601 meters.
03

Calculate the Horizontal Distance

The horizontal distance (\(d\)) can be found using the formula: \[ d = v_{x} \cdot t \] where \(v_{x} = 1.10 \ \mathrm{m/s}\). \[ d = 1.10 \times 0.350 = 0.385 \ \mathrm{m} \] The horizontal distance is 0.385 meters.
04

Calculate the Velocities Before Impact

Horizontal velocity \(v_{x}\) remains constant at 1.10 \(\mathrm{m/s}\). Vertical velocity \(v_{y}\) just before impact is calculated using: \[ v_{y} = g \cdot t = 9.81 \times 0.350 = 3.434 \ \mathrm{m/s} \] Hence, \(v_{y} = 3.434 \ \mathrm{m/s}\) just before impact.
05

Calculate Magnitude and Direction of Total Velocity

The magnitude of the total velocity \(v\) is calculated by combining horizontal and vertical components: \[ v = \sqrt{v_{x}^2 + v_{y}^2} \]\[ v = \sqrt{(1.10)^2 + (3.434)^2} = 3.61 \ \mathrm{m/s} \] The direction \(\theta\) relative to the horizontal is given by:\[ \theta = \tan^{-1}\left(\frac{v_{y}}{v_{x}}\right) = \tan^{-1}\left(\frac{3.434}{1.10}\right) = 72.3^\circ \] The magnitude is 3.61 m/s and the direction is 72.3 degrees below the horizontal.
06

Draw Motion Graphs

- **Position vs. Time (\(x-t\)):** A constant linear graph depicting \(x = 1.10 t\). - **Position vs. Time (\(y-t\)):** A quadratic curve showing \(y = \frac{1}{2} g t^2\), starting at \(h\) and decreasing until hitting zero.- **Velocity vs. Time (\(v_{x}-t\)):** A horizontal line at \(1.10 \ \mathrm{m/s}\).- **Velocity vs. Time (\(v_{y}-t\)):** A linear increase from zero, showing \(v_{y} = g t\).

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
When an object is dropped or moves downwards only under the influence of gravity, it is said to be in free fall. In the case of the physics book sliding off the tabletop, the book undergoes free fall once it leaves the table's edge.
This means its motion after leaving the table is solely affected by gravity, which acts to accelerate it downwards.
Key points about free fall:
  • The only force acting is gravity, with no air resistance.
  • In a vacuum, all objects fall at the same rate regardless of their mass.
  • The acceleration due to gravity is typically given as 9.81 m/s extsuperscript{2} on Earth.
  • The downward velocity increases linearly with time, described by the equation: \( v = g imes t \).
This free-fall motion allows us to calculate important parameters like the fall height using the equation \( h = \frac{1}{2} g t^2 \), as seen in the exercise solution. Understanding free fall is essential to predict how long and how far objects will fall.
Horizontal Velocity
Horizontal velocity is the component of velocity parallel to the horizontal surface. As in the scenario described, when the physics book slides off the tabletop, it already has a given horizontal velocity of 1.10 m/s.
Horizontal velocity is unaffected by gravity. It remains constant as there are no horizontal forces (like air resistance, in this idealized situation) acting on the object.Important considerations for horizontal velocity:
  • Constant over time unless acted upon by an external force.
  • Described using \( v_{x} = \) initial horizontal velocity, which is 1.10 m/s for the book.
  • Distance covered horizontally is calculated by \( d = v_{x} imes t \).
Thus in our case, the constant horizontal motion helps calculate how far the book travels before it hits the ground. Its consistent nature is perfect for predicting the projectile's horizontal path.
Vertical Velocity
Vertical velocity changes as the object is influenced by gravity. Once the book leaves the table, its vertical velocity starts at zero and increases as gravity acts on it.
Vertical velocity at any point can be calculated using the formula \( v_{y} = g imes t \), where \( g \) is 9.81 m/s extsuperscript{2} and \( t \) is the time since the onset of the free fall.Key features of vertical velocity:
  • Changes at a rate of 9.81 m/s extsuperscript{2} due to gravity.
  • Initial vertical velocity is zero if the fall starts from rest.
  • In this exercise, before impact, \( v_{y} \) reaches 3.434 m/s.
Vertical velocity is crucial for understanding how quickly an object will fall and creates a meaningful way to assess motion as objects move toward the ground in projectile events.
Motion Graphs
Motion graphs visually present how an object's position and velocity change over time, helping to illustrate the motion characteristics.
**Position vs. Time Graphs**:
  • For horizontal position (\(x-t\)), it shows a straight line, indicating constant speed, reflecting the consistent horizontal velocity.
  • For vertical position (\(y-t\)), it appears as a parabola, showing the acceleration due to gravity and how the vertical position changes as time progresses.
**Velocity vs. Time Graphs**:
  • The horizontal velocity (\(v_{x}-t\)) is shown as a flat line, emphasizing the constant horizontal speed of 1.10 m/s.
  • The vertical velocity (\(v_{y}-t\)) shows a linear increase, mirroring gravity's constant acceleration effect.
These graphs are fundamental to analyzing motion because they offer a visual and intuitive way to understand detailed dynamics in projectile motion, helping bridge the gap between abstract equations and real-world motion.

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

Two tanks are engaged in a training exercise on level ground. The first tank fires a paint-filled training round with a muzzle speed of 250 \(\mathrm{m} / \mathrm{s}\) at \(10.0^{\circ}\) above the horizontal while advancing toward the second tank with a speed of 15.0 \(\mathrm{m} / \mathrm{s}\) relative to the ground. The second tank is retreating at 35.0 \(\mathrm{m} / \mathrm{s}\) relative to the ground, but is hit by the shell. You can ignore air resistance and assume the shell hits at the same height above ground from which it was fired. Find the distance between the tanks (a) when the round was first fired and (b) at the time of impact.

A railroad flatear is traveling to the right at a speed of 13.0 \(\mathrm{m} / \mathrm{s}\) relative to an observer standing on the ground. Someone is riding a motor scooter on the flatcar (Fig. 3.43\()\) . What is the velocity (magnitude and direction) of the motor scooter relative to the flatcar if its velocity relative to the observer on the ground is (a) 18.0 \(\mathrm{m} / \mathrm{s}\) to the right? (b) 3.0 \(\mathrm{m} / \mathrm{s}\) to the left? (c) zero?

A rock tied to a rope moves in the \(x y\) -plane. Its coordinates are given as functions of time by $$ x(t)=R \cos \omega t \quad y(t)=R \sin \omega t $$ where \(R\) and \(\omega\) are constants. (a) Show that the rock's distance from the origin is constant and equal to \(R-\) that is, that its path is a circle of radius \(R .(b)\) Show that at every point the rock's velocity is perpendicular to its position vector. (c) Show that the rock's acceleration is always opposite in direction to its position vector and has magnitude \(\omega^{2} R .\) (d) Show that the magnitude of the rock's velocity is constant and equal to \(\omega R\) . (e) Combine the results of parts \((\mathrm{c})\) and \((\mathrm{d})\) to show that the rock's acceleration has constant magnitude \(v^{2} / R .\)

An airlane pilit wishes to fy due west. A wind of 80.0 \(\mathrm{km} / \mathrm{h}\) (about 50 \(\mathrm{mi} / \mathrm{h} )\) is blowing toward the south. (a) If the airspeed of the plane (its speed in still air) is 320.0 \(\mathrm{km} / \mathrm{h}\) (about 200 \(\mathrm{mi} / \mathrm{h} )\) , in which direction should the pilot head? (b) What is the speed of the plane over the ground? TIlustrate with a vector diagram.

Crossing the River I. A river flows due south with a specd of 2.0 \(\mathrm{m} / \mathrm{s}\) . A man steers a motorboat across the river, his velocity relative to the water is 4.2 \(\mathrm{m} / \mathrm{s}\) due east. The river is 800 \(\mathrm{m}\) wide. (a) What is his velocity (magnitude and direction) relative to the earth? (b) How much time is required to cross the river? (c) How far south of his starting point will he reach the opposite bank?
See all solutions

Recommended explanations on Physics Textbooks

Physics of Motion

Read Explanation

Engineering Physics

Read Explanation

Classical Mechanics

Read Explanation

Kinematics Physics

Read Explanation

Measurements

Read Explanation

Fundamentals of Physics

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.