### Physics Problem: Calculating the Height and Velocity **Problem Statement:**1. A car light is thrown downward from the top of the Empire State Building with an initial speed of 2.0 meters per second.2. The light takes 4.50 seconds to reach the ground.3. Ignore air resistance. **Tasks:**1. Find the height of the Empire State Building.2. Find the magnitude of the light’s velocity before it reaches the ground.Let's break down how to solve this problem step by step.#### Finding the height of the Empire State BuildingTo find the height (h), we use the equations of motion. The acceleration is due to gravity (g = 9.8 m/s²), the initial velocity (u) is 2.0 m/s, and the time (t) is 4.50 seconds.The formula we use is:\[ h = ut + \frac{1}{2}gt^2 \]Substitute the given values:\[ h = 2.0 \, \text{m/s} \times 4.50 \, \text{s} + \frac{1}{2} \times 9.8 \, \text{m/s}^2 \times (4.50 \, \text{s})^2 \]\[ h = 9 \, \text{m} + \frac{1}{2} \times 9.8 \, \text{m/s}^2 \times 20.25 \, \text{s}^2 \]\[ h = 9 \, \text{m} + 99.225 \, \text{m} \]\[ h = 108.225 \, \text{m} \]**Height of the Empire State Building: 108.225 meters**#### Finding the magnitude of the light's velocity before it reaches the groundTo find the final velocity (v), we again use the equations of motion. We know initial velocity (u), acceleration (a), and time (t):\[ v = u + gt \]Substitute the given values:\[ v = 2.0 \, \text{m/s} + 9.8 \, \text{m/s}^2 \times 4.50 \, \text{s} \]\[ v = 2.0 \, \text{m

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14. A car light is thrown downward from the top of the empire state building with an initial speed 2.0 meters per second. 4.50 seconds is the time it takes for the light to hit the ground. Ignore air resistance. Find the height of the empire state building. Find the magnitude of the light's velocity before it reaches the ground.

14. A car light is thrown downward from the top of the empire state building with an initial speed 2.0 meters per second. 4.50 seconds is the time it takes for the light to hit the ground. Ignore air resistance. Find the height of the empire state building. Find the magnitude of the light's velocity before it reaches the ground.

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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Concept explainers

Displacement, Velocity and Acceleration

In classical mechanics, kinematics deals with the motion of a particle. It deals only with the position, velocity, acceleration, and displacement of a particle. It has no concern about the source of motion.

Linear Displacement

The term "displacement" refers to when something shifts away from its original "location," and "linear" refers to a straight line. As a result, “Linear Displacement” can be described as the movement of an object in a straight line along a single axis, for example, from side to side or up and down. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Linear displacement is usually measured in millimeters or inches and may be positive or negative.

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Question

### Physics Problem: Calculating the Height and Velocity

**Problem Statement:**

1. A car light is thrown downward from the top of the Empire State Building with an initial speed of 2.0 meters per second.
2. The light takes 4.50 seconds to reach the ground.
3. Ignore air resistance.

**Tasks:**

1. Find the height of the Empire State Building.
2. Find the magnitude of the light’s velocity before it reaches the ground.

Let's break down how to solve this problem step by step.

#### Finding the height of the Empire State Building

To find the height (h), we use the equations of motion. The acceleration is due to gravity (g = 9.8 m/s²), the initial velocity (u) is 2.0 m/s, and the time (t) is 4.50 seconds.

The formula we use is:

\[ h = ut + \frac{1}{2}gt^2 \]

Substitute the given values:

\[ h = 2.0 \, \text{m/s} \times 4.50 \, \text{s} + \frac{1}{2} \times 9.8 \, \text{m/s}^2 \times (4.50 \, \text{s})^2 \]

\[ h = 9 \, \text{m} + \frac{1}{2} \times 9.8 \, \text{m/s}^2 \times 20.25 \, \text{s}^2 \]

\[ h = 9 \, \text{m} + 99.225 \, \text{m} \]

\[ h = 108.225 \, \text{m} \]

**Height of the Empire State Building: 108.225 meters**

#### Finding the magnitude of the light's velocity before it reaches the ground

To find the final velocity (v), we again use the equations of motion. We know initial velocity (u), acceleration (a), and time (t):

\[ v = u + gt \]

Substitute the given values:

\[ v = 2.0 \, \text{m/s} + 9.8 \, \text{m/s}^2 \times 4.50 \, \text{s} \]

\[ v = 2.0 \, \text{m

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