Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 2: Problem 12

You toss a small ball vertically up in the air. How are the velocity and acceleration vectors of the ball oriented with respect to one another during the ball's flight up and down?

Short Answer

Expert verified
Answer: During the upward flight, the velocity vector points upward and the acceleration vector (due to gravity) points downward. At the highest point, the velocity vector is momentarily zero (no direction) while the acceleration vector points downward. During the downward flight, both the velocity and acceleration vectors point downward.

Step by step solution

01

Identify the given information

The exercise tells us that a ball is tossed vertically up in the air. The only acceleration acting on the ball is due to gravity. The acceleration due to gravity has a constant value, and it points downwards.
02

Analyze the upward flight of the ball

During the upward flight, the velocity vector of the ball is pointing upward since the ball is moving up. However, the acceleration due to gravity is still acting downward. So during this phase, the two vectors have opposite directions.
03

Analyze the highest point of the flight

At the highest point of the ball's flight, the velocity vector becomes momentarily zero since the ball is neither moving up nor down. The acceleration due to gravity remains constant and points downward. At this point, the velocity vector can be regarded as having no direction, while the acceleration vector has a downward orientation.
04

Analyze the downward flight of the ball

During the downward flight, the velocity vector of the ball is pointing downward since the ball is moving in that direction. At the same time, the acceleration due to gravity continues to act downward. In this phase, the two vectors have the same direction.
05

Summarize the orientation of velocity and acceleration vectors

To answer the exercise, we have analyzed the orientation of the velocity and acceleration vectors throughout the ball's flight: 1. During the upward flight, the velocity vector points upward while the acceleration due to gravity points downward. They have opposite directions. 2. At the highest point, the velocity vector is momentarily zero (no direction) while the acceleration due to gravity is still downward. 3. During the downward flight, both the velocity and acceleration vectors point downward. They have the same direction.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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 car is traveling due west at \(20.0 \mathrm{~m} / \mathrm{s}\). Find the velocity of the car after \(3.00 \mathrm{~s}\) if its acceleration is \(1.0 \mathrm{~m} / \mathrm{s}^{2}\) due west. Assume the acceleration remains constant. a) \(17.0 \mathrm{~m} / \mathrm{s}\) west b) \(17.0 \mathrm{~m} / \mathrm{s}\) east c) \(23.0 \mathrm{~m} / \mathrm{s}\) west d) \(23.0 \mathrm{~m} / \mathrm{s}\) east e) \(11.0 \mathrm{~m} / \mathrm{s}\) south

A runner of mass 57.5 kg starts from rest and accelerates with a constant acceleration of \(1.25 \mathrm{~m} / \mathrm{s}^{2}\) until she reaches a velocity of \(6.3 \mathrm{~m} / \mathrm{s}\). She then continues running with this constant velocity. a) How far has she run after 59.7 s? b) What is the velocity of the runner at this point?

You are driving at \(29.1 \mathrm{~m} / \mathrm{s}\) when the truck ahead of you comes to a halt \(200.0 \mathrm{~m}\) away from your bumper. Your brakes are in poor condition and you decelerate at a constant rate of \(2.4 \mathrm{~m} / \mathrm{s}^{2}\) a) How close do you come to the bumper of the truck? b) How long does it take you to come to a stop?

The position of a particle moving along the \(x\) -axis is given by \(x=\left(11+14 t-2.0 t^{2}\right),\) where \(t\) is in seconds and \(x\) is in meters. What is the average velocity during the time interval from \(t=1.0 \mathrm{~s}\) to \(t=4.0 \mathrm{~s} ?\)

The trajectory of an object is given by the equation $$ x(t)=(4.35 \mathrm{~m})+(25.9 \mathrm{~m} / \mathrm{s}) t-\left(11.79 \mathrm{~m} / \mathrm{s}^{2}\right) t^{2} $$ a) For which time \(t\) is the displacement \(x(t)\) at its maximum? b) What is this maximum value?
See all solutions

Recommended explanations on Physics Textbooks

Fluids

Read Explanation

Dynamics

Read Explanation

Physical Quantities And Units

Read Explanation

Kinematics Physics

Read Explanation

Particle Model of Matter

Read Explanation

Work Energy and Power

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.

Sign-up for free