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College Physics by Openstax

You can browse on the itemized questions with solutions of the College Physics by Openstax below. Also, you can buy the whole Complete Solution Manual here.

College Physics Cover of Chapter 1: Introduction: The Nature of Science and Physics

Chapter 1: Introduction: The Nature of Science and Physics

College Physics Cover of Chapter 2: Kinematics

Chapter 2: Kinematics

College Physics Cover of Chapter 3: Two-Dimensional Kinematics

Chapter 3: Two-Dimensional Kinematics

College Physics Cover of Chapter 4 Dynamics: Force and Newton's Law of Motion

Chapter 4: Dynamics: Force and Newton’s Law of Motion

College Physics Cover of Chapter 5 Further Applications of Newton's Laws: Friction, Drag, and Elasticity

Chapter 5: Further Applications of Newton’s Laws: Friction, Drag, and Elasticity

College Physics Cover of Chapter 6 Uniform Circular Motion and Gravitation

Chapter 6: Uniform Circular Motion and Gravitation

College Physics Cover of Chapter 7 Work, Energy, and Energy Resources

Chapter 7: Work, Energy, and Energy Resources

College Physics Cover of Chapter 8 Linear Momentum and Collisions

Chapter 8: Linear Momentum and Collisions

College Physics Cover of Chapter 9

Chapter 9: Statics and Torque

College Physics Cover of Chapter 10

Chapter 10: Rotational Motion and Angular Momentum

College Physics Cover of Chapter 11

Chapter 11: Fluid Mechanics

College Physics Cover of Chapter 12

Chapter 12: Fluid Dynamics and Its Biological and Medical Applications

College Physics Cover of Chapter 13

Chapter 13: Temperature, Kinetic Theory, and the Gas Laws

College Physics Cover of Chapter 14

Chapter 14: Heat and Heat Transfer Methods

College Physics Cover of Chapter 15

Chapter 15: Thermodynamics

College Physics Cover of Chapter 16

Chapter 16: Oscillatory Motion and Waves

College Physics Cover of Chapter 17

Chapter 17: Physics of Hearing

College Physics Cover of Chapter 18

Chapter 18:
Electric Charge and Electric Field

College Physics Cover of Chapter 19

Chapter 19:
Electric Potential and Electric Field

College Physics Cover of Chapter 20

Chapter 20:
Electric Current, Resistance, and Ohm’s Law

College Physics Cover of Chapter 21

Chapter 21: Circuits and DC Instruments

College Physics Cover of Chapter 22

Chapter 22:
Magnetism

College Physics Cover of Chapter 23

Chapter 23:
Electromagnetic Induction, AC Circuits, and Electrical Technologies

College Physics Cover of Chapter 24

Chapter 24:
Electromagnetic Waves

College Physics Cover of Chapter 25

Chapter 25: Geometric Optics

College Physics Cover of Chapter 26

Chapter 26: Vision and Optical Instrument

College Physics Cover of Chapter 27

Chapter 27: Wave Optics

College Physics Cover of Chapter 28

Chapter 28: Special Relativity

College Physics Cover of Chapter 29

Chapter 29: Introduction to Quantum Physics

College Physics Cover of Chapter 30

Chapter 30: Atomic Physics

College Physics Cover of Chapter 31

Chapter 31: Radioactivity and Nuclear Physics

College Physics Cover of Chapter 32

Chapter 32: Medical Applications of Nuclear Physics

College Physics Cover of Chapter 33

Chapter 33:
Particle Physics

College Physics Cover of Chapter 34

Chapter 34: Frontiers of Physics

College Physics by Openstax Chapter 2 Problem 48

A very strong, but inept, shot putter puts the shot straight up vertically with an initial velocity of 11.0 m/s. How long does he have to get out of the way if the shot was released at a height of 2.20 m, and he is 1.80 m tall?

Solution:

The known values are: $y_0=2.20\:\text{m}$ ; $y=1.80\:\text{m}$ ; $v_0=11.0\:\text{m/s}$ ; and $a=-9.80\:\text{m/s}^2$

We are going to use the formula

\Delta y=v_0t+\frac{1}{2}at^2

Substituting the given values:

\begin{align*} \Delta y & =v_0t+\frac{1}{2}at^2 \\ 1.80\:\text{m}-2.20\:\text{m} & =\left(11.0\:\text{m/s}\right)t+\frac{1}{2}\left(-9.80\:\text{m/s}^2\right)t^2 \\ -0.40\:\text{m} & =\left(11.0\:\text{m/s}\right)t-\left(4.90\:\text{m/s}^2\right)t^2 \\ 4.90t^2-11t-0.40 & =0 \end{align*}

Using the quadratic formula solve for $t$ , we have

\begin{align*} t & =\frac{-\left(-11\right)\pm \sqrt{\left(-11\right)^2-4\left(4.90\right)\left(-0.40\right)}}{2\left(4.90\right)} \\ \end{align*}

t=2.28\:\text{sec}\:\:\:\:\:\text{or}\:\:\:\:\:\:t=-0.04 \ \text{sec}

We can discard the negative time, so

t=2.28\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

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