# A. An object is dropped from a height of 20 feet off the ground. What is its velocity when it hits the ground? B. Instead of being dropped, the object is thrown down, such that when it is 20 feet off the ground, it already has an initial velocity of 20 ft/s straight down. What is its velocity when it hits the ground? C. What did you assume in answering parts A and B? Give at least three examples of objects for which your assumptions are very good, and at least one example of an object for which your assumptions would fail badly.

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### Fundamentals of Chemical Engineeri...

1st Edition
Kevin D. Dahm + 1 other
Publisher: Cengage Learning
ISBN: 9781111580704
BuyFind

### Fundamentals of Chemical Engineeri...

1st Edition
Kevin D. Dahm + 1 other
Publisher: Cengage Learning
ISBN: 9781111580704

#### Solutions

Chapter
Section
Chapter 1.7, Problem 22P
Textbook Problem

## A. An object is dropped from a height of 20 feet off the ground. What is its velocity when it hits the ground?B. Instead of being dropped, the object is thrown down, such that when it is 20 feet off the ground, it already has an initial velocity of 20 ft/s straight down. What is its velocity when it hits the ground?C. What did you assume in answering parts A and B? Give at least three examples of objects for which your assumptions are very good, and at least one example of an object for which your assumptions would fail badly.

Expert Solution

A)

Interpretation Introduction

Interpretation:

The velocity of dropped object when hits the ground

Concept introduction:

Law of conservation of energy:

The energy can neither be created nor be destroyed but it changes from one form to another. Accordingly, the energy entering into the open system will be equivalent to the energy leaving out of the same system, thereby conserving energy.

ΔP.E=ΔK.EΔP.E+ΔK.E=0MgΔz+MΔv22=0(v22v12)2=g(z2z1)v22=v12+[2g(z2z1)]

Here, gravity is g, mass is M, velocity is v, initial height is z1, final height is z2, and difference in height is Δz.

### Explanation of Solution

Given information:

The initial height of the object =20 ft

Acceleration due to gravity =32.2ft/s2

Calculate the velocity of an object hits the ground.

v22=v12+[2g(z2z1)]        (1)

Substitute 32.2ft/s2 for g, 20 ft for z1, 0 for v1 and 0 for z2 in equation (1)

Expert Solution

B)

Interpretation Introduction

Interpretation:

The velocity of thrown down object when hits the ground

Concept introduction:

Law of conservation of energy:

The energy can neither be created nor be destroyed but it changes from one form to another. Accordingly, the energy entering into the open system will be equivalent to the energy leaving out of the same system, thereby conserving energy.

The law of energy conservation for the thrown ball

ΔP.E+ΔK.E=0MgΔz+MΔv22=0(v22v12)2=g(z2z1)v2=2g(z1z2)+v12

Expert Solution

C)

Interpretation Introduction

Interpretation:

Examples for the assumption made in part A and B have to be determined.

Concept introduction:

Air resistance is depending on the ratio of surface area to the mass, if the mass is less and the surface area is more means air resistance is more.

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