**Problem 2**: What is the displacement of a ball that is thrown up at 14.0 m/s one second after being thrown?To understand the problem, we can apply physics principles, particularly the equations of motion under constant acceleration due to gravity (approximately \(-9.8 \, \text{m/s}^2\) downward).We will need to calculate the displacement (change in position) of the ball one second after it is thrown. The initial velocity (\(u\)) is \(14.0 \, \text{m/s}\) upward, and time (\(t\)) is \(1 \, \text{s}\).The formula for displacement (\(s\)) is:\[ s = ut + \frac{1}{2} a t^2 \]Where:- \(u\) is the initial velocity,- \(a\) is the acceleration due to gravity (use \(-9.8 \, \text{m/s}^2\)),- \(t\) is the time.Substitute the known values:\[ s = 14.0 \, \text{m/s} \times 1 \, \text{s} + \frac{1}{2} \times (-9.8 \, \text{m/s}^2) \times (1 \, \text{s})^2 \]\[ s = 14.0 \, \text{m} - 4.9 \, \text{m} \]\[ s = 9.1 \, \text{m} \]Therefore, the displacement of the ball after one second is \(9.1 \, \text{m}\) upward.

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2). What is the displacement of a ball that is thrown up at 14.0 m/s one second after thrown?

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Question

**Problem 2**: What is the displacement of a ball that is thrown up at 14.0 m/s one second after being thrown?

To understand the problem, we can apply physics principles, particularly the equations of motion under constant acceleration due to gravity (approximately \(-9.8 \, \text{m/s}^2\) downward).

We will need to calculate the displacement (change in position) of the ball one second after it is thrown. The initial velocity (\(u\)) is \(14.0 \, \text{m/s}\) upward, and time (\(t\)) is \(1 \, \text{s}\).

The formula for displacement (\(s\)) is:

\[ s = ut + \frac{1}{2} a t^2 \]

Where:
- \(u\) is the initial velocity,
- \(a\) is the acceleration due to gravity (use \(-9.8 \, \text{m/s}^2\)),
- \(t\) is the time.

Substitute the known values:

\[ s = 14.0 \, \text{m/s} \times 1 \, \text{s} + \frac{1}{2} \times (-9.8 \, \text{m/s}^2) \times (1 \, \text{s})^2 \]

\[ s = 14.0 \, \text{m} - 4.9 \, \text{m} \]

\[ s = 9.1 \, \text{m} \]

Therefore, the displacement of the ball after one second is \(9.1 \, \text{m}\) upward.

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