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  1. Preface
  2. 1 Introduction: The Nature of Science and Physics
    1. Connection for AP® Courses
    2. 1.1 Physics: An Introduction
    3. 1.2 Physical Quantities and Units
    4. 1.3 Accuracy, Precision, and Significant Figures
    5. 1.4 Approximation
    6. Glossary
    7. Section Summary
    8. Conceptual Questions
    9. Problems & Exercises
  3. 2 Kinematics
    1. Connection for AP® Courses
    2. 2.1 Displacement
    3. 2.2 Vectors, Scalars, and Coordinate Systems
    4. 2.3 Time, Velocity, and Speed
    5. 2.4 Acceleration
    6. 2.5 Motion Equations for Constant Acceleration in One Dimension
    7. 2.6 Problem-Solving Basics for One Dimensional Kinematics
    8. 2.7 Falling Objects
    9. 2.8 Graphical Analysis of One Dimensional Motion
    10. Glossary
    11. Section Summary
    12. Conceptual Questions
    13. Problems & Exercises
    14. Test Prep for AP® Courses
  4. 3 Two-Dimensional Kinematics
    1. Connection for AP® Courses
    2. 3.1 Kinematics in Two Dimensions: An Introduction
    3. 3.2 Vector Addition and Subtraction: Graphical Methods
    4. 3.3 Vector Addition and Subtraction: Analytical Methods
    5. 3.4 Projectile Motion
    6. 3.5 Addition of Velocities
    7. Glossary
    8. Section Summary
    9. Conceptual Questions
    10. Problems & Exercises
    11. Test Prep for AP® Courses
  5. 4 Dynamics: Force and Newton's Laws of Motion
    1. Connection for AP® Courses
    2. 4.1 Development of Force Concept
    3. 4.2 Newton's First Law of Motion: Inertia
    4. 4.3 Newton's Second Law of Motion: Concept of a System
    5. 4.4 Newton's Third Law of Motion: Symmetry in Forces
    6. 4.5 Normal, Tension, and Other Examples of Force
    7. 4.6 Problem-Solving Strategies
    8. 4.7 Further Applications of Newton's Laws of Motion
    9. 4.8 Extended Topic: The Four Basic Forces—An Introduction
    10. Glossary
    11. Section Summary
    12. Conceptual Questions
    13. Problems & Exercises
    14. Test Prep for AP® Courses
  6. 5 Further Applications of Newton's Laws: Friction, Drag, and Elasticity
    1. Connection for AP® Courses
    2. 5.1 Friction
    3. 5.2 Drag Forces
    4. 5.3 Elasticity: Stress and Strain
    5. Glossary
    6. Section Summary
    7. Conceptual Questions
    8. Problems & Exercises
    9. Test Prep for AP® Courses
  7. 6 Gravitation and Uniform Circular Motion
    1. Connection for AP® Courses
    2. 6.1 Rotation Angle and Angular Velocity
    3. 6.2 Centripetal Acceleration
    4. 6.3 Centripetal Force
    5. 6.4 Fictitious Forces and Non-inertial Frames: The Coriolis Force
    6. 6.5 Newton's Universal Law of Gravitation
    7. 6.6 Satellites and Kepler's Laws: An Argument for Simplicity
    8. Glossary
    9. Section Summary
    10. Conceptual Questions
    11. Problems & Exercises
    12. Test Prep for AP® Courses
  8. 7 Work, Energy, and Energy Resources
    1. Connection for AP® Courses
    2. 7.1 Work: The Scientific Definition
    3. 7.2 Kinetic Energy and the Work-Energy Theorem
    4. 7.3 Gravitational Potential Energy
    5. 7.4 Conservative Forces and Potential Energy
    6. 7.5 Nonconservative Forces
    7. 7.6 Conservation of Energy
    8. 7.7 Power
    9. 7.8 Work, Energy, and Power in Humans
    10. 7.9 World Energy Use
    11. Glossary
    12. Section Summary
    13. Conceptual Questions
    14. Problems & Exercises
    15. Test Prep for AP® Courses
  9. 8 Linear Momentum and Collisions
    1. Connection for AP® courses
    2. 8.1 Linear Momentum and Force
    3. 8.2 Impulse
    4. 8.3 Conservation of Momentum
    5. 8.4 Elastic Collisions in One Dimension
    6. 8.5 Inelastic Collisions in One Dimension
    7. 8.6 Collisions of Point Masses in Two Dimensions
    8. 8.7 Introduction to Rocket Propulsion
    9. Glossary
    10. Section Summary
    11. Conceptual Questions
    12. Problems & Exercises
    13. Test Prep for AP® Courses
  10. 9 Statics and Torque
    1. Connection for AP® Courses
    2. 9.1 The First Condition for Equilibrium
    3. 9.2 The Second Condition for Equilibrium
    4. 9.3 Stability
    5. 9.4 Applications of Statics, Including Problem-Solving Strategies
    6. 9.5 Simple Machines
    7. 9.6 Forces and Torques in Muscles and Joints
    8. Glossary
    9. Section Summary
    10. Conceptual Questions
    11. Problems & Exercises
    12. Test Prep for AP® Courses
  11. 10 Rotational Motion and Angular Momentum
    1. Connection for AP® Courses
    2. 10.1 Angular Acceleration
    3. 10.2 Kinematics of Rotational Motion
    4. 10.3 Dynamics of Rotational Motion: Rotational Inertia
    5. 10.4 Rotational Kinetic Energy: Work and Energy Revisited
    6. 10.5 Angular Momentum and Its Conservation
    7. 10.6 Collisions of Extended Bodies in Two Dimensions
    8. 10.7 Gyroscopic Effects: Vector Aspects of Angular Momentum
    9. Glossary
    10. Section Summary
    11. Conceptual Questions
    12. Problems & Exercises
    13. Test Prep for AP® Courses
  12. 11 Fluid Statics
    1. Connection for AP® Courses
    2. 11.1 What Is a Fluid?
    3. 11.2 Density
    4. 11.3 Pressure
    5. 11.4 Variation of Pressure with Depth in a Fluid
    6. 11.5 Pascal’s Principle
    7. 11.6 Gauge Pressure, Absolute Pressure, and Pressure Measurement
    8. 11.7 Archimedes’ Principle
    9. 11.8 Cohesion and Adhesion in Liquids: Surface Tension and Capillary Action
    10. 11.9 Pressures in the Body
    11. Glossary
    12. Section Summary
    13. Conceptual Questions
    14. Problems & Exercises
    15. Test Prep for AP® Courses
  13. 12 Fluid Dynamics and Its Biological and Medical Applications
    1. Connection for AP® Courses
    2. 12.1 Flow Rate and Its Relation to Velocity
    3. 12.2 Bernoulli’s Equation
    4. 12.3 The Most General Applications of Bernoulli’s Equation
    5. 12.4 Viscosity and Laminar Flow; Poiseuille’s Law
    6. 12.5 The Onset of Turbulence
    7. 12.6 Motion of an Object in a Viscous Fluid
    8. 12.7 Molecular Transport Phenomena: Diffusion, Osmosis, and Related Processes
    9. Glossary
    10. Section Summary
    11. Conceptual Questions
    12. Problems & Exercises
    13. Test Prep for AP® Courses
  14. 13 Temperature, Kinetic Theory, and the Gas Laws
    1. Connection for AP® Courses
    2. 13.1 Temperature
    3. 13.2 Thermal Expansion of Solids and Liquids
    4. 13.3 The Ideal Gas Law
    5. 13.4 Kinetic Theory: Atomic and Molecular Explanation of Pressure and Temperature
    6. 13.5 Phase Changes
    7. 13.6 Humidity, Evaporation, and Boiling
    8. Glossary
    9. Section Summary
    10. Conceptual Questions
    11. Problems & Exercises
    12. Test Prep for AP® Courses
  15. 14 Heat and Heat Transfer Methods
    1. Connection for AP® Courses
    2. 14.1 Heat
    3. 14.2 Temperature Change and Heat Capacity
    4. 14.3 Phase Change and Latent Heat
    5. 14.4 Heat Transfer Methods
    6. 14.5 Conduction
    7. 14.6 Convection
    8. 14.7 Radiation
    9. Glossary
    10. Section Summary
    11. Conceptual Questions
    12. Problems & Exercises
    13. Test Prep for AP® Courses
  16. 15 Thermodynamics
    1. Connection for AP® Courses
    2. 15.1 The First Law of Thermodynamics
    3. 15.2 The First Law of Thermodynamics and Some Simple Processes
    4. 15.3 Introduction to the Second Law of Thermodynamics: Heat Engines and Their Efficiency
    5. 15.4 Carnot’s Perfect Heat Engine: The Second Law of Thermodynamics Restated
    6. 15.5 Applications of Thermodynamics: Heat Pumps and Refrigerators
    7. 15.6 Entropy and the Second Law of Thermodynamics: Disorder and the Unavailability of Energy
    8. 15.7 Statistical Interpretation of Entropy and the Second Law of Thermodynamics: The Underlying Explanation
    9. Glossary
    10. Section Summary
    11. Conceptual Questions
    12. Problems & Exercises
    13. Test Prep for AP® Courses
  17. 16 Oscillatory Motion and Waves
    1. Connection for AP® Courses
    2. 16.1 Hooke’s Law: Stress and Strain Revisited
    3. 16.2 Period and Frequency in Oscillations
    4. 16.3 Simple Harmonic Motion: A Special Periodic Motion
    5. 16.4 The Simple Pendulum
    6. 16.5 Energy and the Simple Harmonic Oscillator
    7. 16.6 Uniform Circular Motion and Simple Harmonic Motion
    8. 16.7 Damped Harmonic Motion
    9. 16.8 Forced Oscillations and Resonance
    10. 16.9 Waves
    11. 16.10 Superposition and Interference
    12. 16.11 Energy in Waves: Intensity
    13. Glossary
    14. Section Summary
    15. Conceptual Questions
    16. Problems & Exercises
    17. Test Prep for AP® Courses
  18. 17 Physics of Hearing
    1. Connection for AP® Courses
    2. 17.1 Sound
    3. 17.2 Speed of Sound, Frequency, and Wavelength
    4. 17.3 Sound Intensity and Sound Level
    5. 17.4 Doppler Effect and Sonic Booms
    6. 17.5 Sound Interference and Resonance: Standing Waves in Air Columns
    7. 17.6 Hearing
    8. 17.7 Ultrasound
    9. Glossary
    10. Section Summary
    11. Conceptual Questions
    12. Problems & Exercises
    13. Test Prep for AP® Courses
  19. 18 Electric Charge and Electric Field
    1. Connection for AP® Courses
    2. 18.1 Static Electricity and Charge: Conservation of Charge
    3. 18.2 Conductors and Insulators
    4. 18.3 Conductors and Electric Fields in Static Equilibrium
    5. 18.4 Coulomb’s Law
    6. 18.5 Electric Field: Concept of a Field Revisited
    7. 18.6 Electric Field Lines: Multiple Charges
    8. 18.7 Electric Forces in Biology
    9. 18.8 Applications of Electrostatics
    10. Glossary
    11. Section Summary
    12. Conceptual Questions
    13. Problems & Exercises
    14. Test Prep for AP® Courses
  20. 19 Electric Potential and Electric Field
    1. Connection for AP® Courses
    2. 19.1 Electric Potential Energy: Potential Difference
    3. 19.2 Electric Potential in a Uniform Electric Field
    4. 19.3 Electrical Potential Due to a Point Charge
    5. 19.4 Equipotential Lines
    6. 19.5 Capacitors and Dielectrics
    7. 19.6 Capacitors in Series and Parallel
    8. 19.7 Energy Stored in Capacitors
    9. Glossary
    10. Section Summary
    11. Conceptual Questions
    12. Problems & Exercises
    13. Test Prep for AP® Courses
  21. 20 Electric Current, Resistance, and Ohm's Law
    1. Connection for AP® Courses
    2. 20.1 Current
    3. 20.2 Ohm’s Law: Resistance and Simple Circuits
    4. 20.3 Resistance and Resistivity
    5. 20.4 Electric Power and Energy
    6. 20.5 Alternating Current versus Direct Current
    7. 20.6 Electric Hazards and the Human Body
    8. 20.7 Nerve Conduction–Electrocardiograms
    9. Glossary
    10. Section Summary
    11. Conceptual Questions
    12. Problems & Exercises
    13. Test Prep for AP® Courses
  22. 21 Circuits, Bioelectricity, and DC Instruments
    1. Connection for AP® Courses
    2. 21.1 Resistors in Series and Parallel
    3. 21.2 Electromotive Force: Terminal Voltage
    4. 21.3 Kirchhoff’s Rules
    5. 21.4 DC Voltmeters and Ammeters
    6. 21.5 Null Measurements
    7. 21.6 DC Circuits Containing Resistors and Capacitors
    8. Glossary
    9. Section Summary
    10. Conceptual Questions
    11. Problems & Exercises
    12. Test Prep for AP® Courses
  23. 22 Magnetism
    1. Connection for AP® Courses
    2. 22.1 Magnets
    3. 22.2 Ferromagnets and Electromagnets
    4. 22.3 Magnetic Fields and Magnetic Field Lines
    5. 22.4 Magnetic Field Strength: Force on a Moving Charge in a Magnetic Field
    6. 22.5 Force on a Moving Charge in a Magnetic Field: Examples and Applications
    7. 22.6 The Hall Effect
    8. 22.7 Magnetic Force on a Current-Carrying Conductor
    9. 22.8 Torque on a Current Loop: Motors and Meters
    10. 22.9 Magnetic Fields Produced by Currents: Ampere’s Law
    11. 22.10 Magnetic Force between Two Parallel Conductors
    12. 22.11 More Applications of Magnetism
    13. Glossary
    14. Section Summary
    15. Conceptual Questions
    16. Problems & Exercises
    17. Test Prep for AP® Courses
  24. 23 Electromagnetic Induction, AC Circuits, and Electrical Technologies
    1. Connection for AP® Courses
    2. 23.1 Induced Emf and Magnetic Flux
    3. 23.2 Faraday’s Law of Induction: Lenz’s Law
    4. 23.3 Motional Emf
    5. 23.4 Eddy Currents and Magnetic Damping
    6. 23.5 Electric Generators
    7. 23.6 Back Emf
    8. 23.7 Transformers
    9. 23.8 Electrical Safety: Systems and Devices
    10. 23.9 Inductance
    11. 23.10 RL Circuits
    12. 23.11 Reactance, Inductive and Capacitive
    13. 23.12 RLC Series AC Circuits
    14. Glossary
    15. Section Summary
    16. Conceptual Questions
    17. Problems & Exercises
    18. Test Prep for AP® Courses
  25. 24 Electromagnetic Waves
    1. Connection for AP® Courses
    2. 24.1 Maxwell’s Equations: Electromagnetic Waves Predicted and Observed
    3. 24.2 Production of Electromagnetic Waves
    4. 24.3 The Electromagnetic Spectrum
    5. 24.4 Energy in Electromagnetic Waves
    6. Glossary
    7. Section Summary
    8. Conceptual Questions
    9. Problems & Exercises
    10. Test Prep for AP® Courses
  26. 25 Geometric Optics
    1. Connection for AP® Courses
    2. 25.1 The Ray Aspect of Light
    3. 25.2 The Law of Reflection
    4. 25.3 The Law of Refraction
    5. 25.4 Total Internal Reflection
    6. 25.5 Dispersion: The Rainbow and Prisms
    7. 25.6 Image Formation by Lenses
    8. 25.7 Image Formation by Mirrors
    9. Glossary
    10. Section Summary
    11. Conceptual Questions
    12. Problems & Exercises
    13. Test Prep for AP® Courses
  27. 26 Vision and Optical Instruments
    1. Connection for AP® Courses
    2. 26.1 Physics of the Eye
    3. 26.2 Vision Correction
    4. 26.3 Color and Color Vision
    5. 26.4 Microscopes
    6. 26.5 Telescopes
    7. 26.6 Aberrations
    8. Glossary
    9. Section Summary
    10. Conceptual Questions
    11. Problems & Exercises
    12. Test Prep for AP® Courses
  28. 27 Wave Optics
    1. Connection for AP® Courses
    2. 27.1 The Wave Aspect of Light: Interference
    3. 27.2 Huygens's Principle: Diffraction
    4. 27.3 Young’s Double Slit Experiment
    5. 27.4 Multiple Slit Diffraction
    6. 27.5 Single Slit Diffraction
    7. 27.6 Limits of Resolution: The Rayleigh Criterion
    8. 27.7 Thin Film Interference
    9. 27.8 Polarization
    10. 27.9 *Extended Topic* Microscopy Enhanced by the Wave Characteristics of Light
    11. Glossary
    12. Section Summary
    13. Conceptual Questions
    14. Problems & Exercises
    15. Test Prep for AP® Courses
  29. 28 Special Relativity
    1. Connection for AP® Courses
    2. 28.1 Einstein’s Postulates
    3. 28.2 Simultaneity And Time Dilation
    4. 28.3 Length Contraction
    5. 28.4 Relativistic Addition of Velocities
    6. 28.5 Relativistic Momentum
    7. 28.6 Relativistic Energy
    8. Glossary
    9. Section Summary
    10. Conceptual Questions
    11. Problems & Exercises
    12. Test Prep for AP® Courses
  30. 29 Introduction to Quantum Physics
    1. Connection for AP® Courses
    2. 29.1 Quantization of Energy
    3. 29.2 The Photoelectric Effect
    4. 29.3 Photon Energies and the Electromagnetic Spectrum
    5. 29.4 Photon Momentum
    6. 29.5 The Particle-Wave Duality
    7. 29.6 The Wave Nature of Matter
    8. 29.7 Probability: The Heisenberg Uncertainty Principle
    9. 29.8 The Particle-Wave Duality Reviewed
    10. Glossary
    11. Section Summary
    12. Conceptual Questions
    13. Problems & Exercises
    14. Test Prep for AP® Courses
  31. 30 Atomic Physics
    1. Connection for AP® Courses
    2. 30.1 Discovery of the Atom
    3. 30.2 Discovery of the Parts of the Atom: Electrons and Nuclei
    4. 30.3 Bohr’s Theory of the Hydrogen Atom
    5. 30.4 X Rays: Atomic Origins and Applications
    6. 30.5 Applications of Atomic Excitations and De-Excitations
    7. 30.6 The Wave Nature of Matter Causes Quantization
    8. 30.7 Patterns in Spectra Reveal More Quantization
    9. 30.8 Quantum Numbers and Rules
    10. 30.9 The Pauli Exclusion Principle
    11. Glossary
    12. Section Summary
    13. Conceptual Questions
    14. Problems & Exercises
    15. Test Prep for AP® Courses
  32. 31 Radioactivity and Nuclear Physics
    1. Connection for AP® Courses
    2. 31.1 Nuclear Radioactivity
    3. 31.2 Radiation Detection and Detectors
    4. 31.3 Substructure of the Nucleus
    5. 31.4 Nuclear Decay and Conservation Laws
    6. 31.5 Half-Life and Activity
    7. 31.6 Binding Energy
    8. 31.7 Tunneling
    9. Glossary
    10. Section Summary
    11. Conceptual Questions
    12. Problems & Exercises
    13. Test Prep for AP® Courses
  33. 32 Medical Applications of Nuclear Physics
    1. Connection for AP® Courses
    2. 32.1 Medical Imaging and Diagnostics
    3. 32.2 Biological Effects of Ionizing Radiation
    4. 32.3 Therapeutic Uses of Ionizing Radiation
    5. 32.4 Food Irradiation
    6. 32.5 Fusion
    7. 32.6 Fission
    8. 32.7 Nuclear Weapons
    9. Glossary
    10. Section Summary
    11. Conceptual Questions
    12. Problems & Exercises
    13. Test Prep for AP® Courses
  34. 33 Particle Physics
    1. Connection for AP® Courses
    2. 33.1 The Yukawa Particle and the Heisenberg Uncertainty Principle Revisited
    3. 33.2 The Four Basic Forces
    4. 33.3 Accelerators Create Matter from Energy
    5. 33.4 Particles, Patterns, and Conservation Laws
    6. 33.5 Quarks: Is That All There Is?
    7. 33.6 GUTs: The Unification of Forces
    8. Glossary
    9. Section Summary
    10. Conceptual Questions
    11. Problems & Exercises
    12. Test Prep for AP® Courses
  35. 34 Frontiers of Physics
    1. Connection for AP® Courses
    2. 34.1 Cosmology and Particle Physics
    3. 34.2 General Relativity and Quantum Gravity
    4. 34.3 Superstrings
    5. 34.4 Dark Matter and Closure
    6. 34.5 Complexity and Chaos
    7. 34.6 High-Temperature Superconductors
    8. 34.7 Some Questions We Know to Ask
    9. Glossary
    10. Section Summary
    11. Conceptual Questions
    12. Problems & Exercises
  36. A | Atomic Masses
  37. B | Selected Radioactive Isotopes
  38. C | Useful Information
  39. D | Glossary of Key Symbols and Notation
  40. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
    8. Chapter 8
    9. Chapter 9
    10. Chapter 10
    11. Chapter 11
    12. Chapter 12
    13. Chapter 13
    14. Chapter 14
    15. Chapter 15
    16. Chapter 16
    17. Chapter 17
    18. Chapter 18
    19. Chapter 19
    20. Chapter 20
    21. Chapter 21
    22. Chapter 22
    23. Chapter 23
    24. Chapter 24
    25. Chapter 25
    26. Chapter 26
    27. Chapter 27
    28. Chapter 28
    29. Chapter 29
    30. Chapter 30
    31. Chapter 31
    32. Chapter 32
    33. Chapter 33
    34. Chapter 34
  41. Index

Learning Objectives

By the end of this section, you will be able to:

  • Identify and explain the properties of a projectile, such as acceleration due to gravity, range, maximum height, and trajectory.
  • Determine the location and velocity of a projectile at different points in its trajectory.
  • Apply the principle of independence of motion to solve projectile motion problems.

The information presented in this section supports the following AP® learning objectives:

  • 3.A.1.1 The student is able to express the motion of an object using narrative, mathematical, and graphical representations. (S.P. 1.5, 2.1, 2.2)
  • 3.A.1.3 The student is able to analyze experimental data describing the motion of an object and is able to express the results of the analysis using narrative, mathematical, and graphical representations. (S.P. 5.1)

Projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. The object is called a projectile, and its path is called its trajectory. The motion of falling objects, as covered in Problem-Solving Basics for One-Dimensional Kinematics, is a simple one-dimensional type of projectile motion in which there is no horizontal movement. In this section, we consider two-dimensional projectile motion, such as that of a football or other object for which air resistance is negligible.

The most important fact to remember here is that motions along perpendicular axes are independent and thus can be analyzed separately. This fact was discussed in Kinematics in Two Dimensions: An Introduction, where vertical and horizontal motions were seen to be independent. The key to analyzing two-dimensional projectile motion is to break it into two motions, one along the horizontal axis and the other along the vertical. (This choice of axes is the most sensible, because acceleration due to gravity is vertical—thus, there will be no acceleration along the horizontal axis when air resistance is negligible.) As is customary, we call the horizontal axis the x-axis and the vertical axis the y-axis. Figure 3.36 illustrates the notation for displacement, where ss size 12{s} {} is defined to be the total displacement and xx size 12{x} {} and yy size 12{y} {} are its components along the horizontal and vertical axes, respectively. The magnitudes of these vectors are s, x, and y. (Note that in the last section we used the notation AA size 12{A} {} to represent a vector with components AxAx size 12{A rSub { size 8{x} } } {} and AyAy size 12{A rSub { size 8{y} } } {}. If we continued this format, we would call displacement ss size 12{s} {} with components sxsx size 12{s rSub { size 8{x} } } {} and sysy size 12{s rSub { size 8{y} } } {}. However, to simplify the notation, we will simply represent the component vectors as xx size 12{x} {} and yy size 12{y} {}.)

Of course, to describe motion we must deal with velocity and acceleration, as well as with displacement. We must find their components along the x- and y-axes, too. We will assume all forces except gravity (such as air resistance and friction, for example) are negligible. The components of acceleration are then very simple: ay=g=9.80 m/s2ay=g=9.80 m/s2 size 12{a rSub { size 8{y} } ="-g"="-9.80" "m/s" rSup { size 8{2} } } {}. (Note that this definition assumes that the upwards direction is defined as the positive direction. If you arrange the coordinate system instead such that the downwards direction is positive, then acceleration due to gravity takes a positive value.) Because gravity is vertical, ax=0ax=0 size 12{a rSub { size 8{x} } } {}. Both accelerations are constant, so the kinematic equations can be used.

Review of Kinematic Equations (constant aa)

x = x 0 + v - t x = x 0 + v - t size 12{x=`x rSub { size 8{0} } `+` { bar {v}}t} {}
3.28
v - = v 0 + v 2 v - = v 0 + v 2 size 12{ { bar {v}}=` { {v rSub { size 8{0} } +v} over {2} } } {}
3.29
v = v 0 + at v = v 0 + at size 12{v=v rSub { size 8{0} } + ital "at"} {}
3.30
x = x 0 + v 0 t + 1 2 at 2 x = x 0 + v 0 t + 1 2 at 2 size 12{x=x rSub { size 8{0} } +v rSub { size 8{0} } t+ { {1} over {2} } ital "at" rSup { size 8{2} } } {}
3.31
v 2 = v 0 2 + 2a ( x x 0 ) . v 2 = v 0 2 + 2a ( x x 0 ) . size 12{v rSup { size 8{2} } =v rSub { size 8{0} } rSup { size 8{2} } +2a \( x - x rSub { size 8{0} } \) } {}
3.32
A soccer player is kicking a soccer ball. The ball travels in a projectile motion and reaches a point whose vertical distance is y and horizontal distance is x. The displacement between the kicking point and the final point is s. The angle made by this displacement vector with x axis is theta.
Figure 3.36 The total displacement ss size 12{s} {} of a soccer ball at a point along its path. The vector ss size 12{s} {} has components xx size 12{x} {} and yy size 12{y} {} along the horizontal and vertical axes. Its magnitude is ss size 12{s} {}, and it makes an angle θθ size 12{θ} {} with the horizontal.

Given these assumptions, the following steps are then used to analyze projectile motion:

Step 1. Resolve or break the motion into horizontal and vertical components along the x- and y-axes. These axes are perpendicular, so Ax=AcosθAx=Acosθ size 12{A rSub { size 8{x} } =A"cos"θ} {} and Ay=AsinθAy=Asinθ size 12{A rSub { size 8{y} } =A"sin"θ} {} are used. The magnitude of the components of displacement ss size 12{s} {} along these axes are xx size 12{x} {} and y.y. size 12{y} {} The magnitudes of the components of the velocity vv size 12{v} {} are vx=vcosθvx=vcosθ size 12{v rSub { size 8{x} } =v"cos"θ} {} and vy=vsinθ,vy=vsinθ, size 12{v rSub { size 8{y} } =v"sin"θ} {} where vv size 12{v} {} is the magnitude of the velocity and θθ size 12{θ} {} is its direction, as shown in Figure 3.37. Initial values are denoted with a subscript 0, as usual.

Step 2. Treat the motion as two independent one-dimensional motions, one horizontal and the other vertical. The kinematic equations for horizontal and vertical motion take the following forms:

Horizontal Motion ( a x = 0 ) Horizontal Motion ( a x = 0 ) size 12{"Horizontal Motion " \( a rSub { size 8{x} } =0 \) } {}
3.33
x = x 0 + v x t x = x 0 + v x t size 12{x=x rSub { size 8{0} } +v rSub { size 8{x} } t} {}
3.34
vx=v0x=vx=velocity is a constant.vx=v0x=vx=velocity is a constant. size 12{v rSub { size 8{x} } =v rSub { size 8{0x} } =v rSub { size 8{x} } ="velocity is a constant."} {}
3.35
Vertical Motion ( assuming positive is up a y = g = 9. 80 m/s 2 ) Vertical Motion ( assuming positive is up a y = g = 9. 80 m/s 2 ) size 12{"Vertical Motion " \( "assuming positive is up "a rSub { size 8{y} } = - g= - 9/"80"" m/s" rSup { size 8{2} } \) } {}
3.36
y = y 0 + 1 2 ( v 0y + v y ) t y = y 0 + 1 2 ( v 0y + v y ) t size 12{y=y rSub { size 8{0} } + { {1} over {2} } \( v rSub { size 8{0y} } +v rSub { size 8{y} } \) t} {}
3.37
v y = v 0y gt v y = v 0y gt size 12{v rSub { size 8{y} } =v rSub { size 8{0y} } - ital "gt"} {}
3.38
y = y 0 + v 0y t 1 2 gt 2 y = y 0 + v 0y t 1 2 gt 2 size 12{y=y rSub { size 8{0} } +v rSub { size 8{0y} } t - { {1} over {2} } ital "gt" rSup { size 8{2} } } {}
3.39
vy2=v 0y 22g(yy0).vy2=v 0y 22g(yy0). size 12{v rSub { size 8{y} } rSup { size 8{2} } =v rSub { size 8{0y} } rSup { size 8{2} } - 2g \( y - y rSub { size 8{0} } \) "."} {}
3.40

Step 3. Solve for the unknowns in the two separate motions—one horizontal and one vertical. Note that the only common variable between the motions is time tt size 12{t} {}. The problem solving procedures here are the same as for one-dimensional kinematics and are illustrated in the solved examples below.

Step 4. Recombine the two motions to find the total displacement ss size 12{s} {} and velocity vv size 12{v} {}. Because the x - and y -motions are perpendicular, we determine these vectors by using the techniques outlined in the Vector Addition and Subtraction: Analytical Methods and employing A = A x 2 + A y 2 A = A x 2 + A y 2 size 12{v= sqrt {v rSub { size 8{x} } rSup { size 8{2} } +v rSub { size 8{y} } rSup { size 8{2} } } } {} and θ=tan1(Ay/Ax)θ=tan1(Ay/Ax) size 12{θ="tan" rSup { size 8{ - 1} } \( A rSub { size 8{y} } /A rSub { size 8{x} } \) } {} in the following form, where θθ size 12{θ} {} is the direction of the displacement ss size 12{s} {} and θvθv size 12{θ rSub { size 8{v} } } {} is the direction of the velocity vv size 12{v} {}:

Total displacement and velocity

s = x 2 + y 2 s = x 2 + y 2 size 12{s= sqrt {x rSup { size 8{2} } +y rSup { size 8{2} } } } {}
3.41
θ = tan 1 ( y / x ) θ = tan 1 ( y / x ) size 12{θ="tan" rSup { size 8{ - 1} } \( y/x \) } {}
3.42
v = v x 2 + v y 2 v = v x 2 + v y 2 size 12{v= sqrt {v rSub { size 8{x} } rSup { size 8{2} } +v rSub { size 8{y} } rSup { size 8{2} } } } {}
3.43
θv=tan1(vy/vx).θv=tan1(vy/vx). size 12{θ rSub { size 8{v} } ="tan" rSup { size 8{ - 1} } \( v rSub { size 8{y} } /v rSub { size 8{x} } \) "."} {}
3.44
In part a the figure shows projectile motion of a ball with initial velocity of v zero at an angle of theta zero with the horizontal x axis. The horizontal component v x and the vertical component v y at various positions of ball in the projectile path is shown. In part b only the horizontal velocity component v sub x is shown whose magnitude is constant at various positions in the path. In part c only vertical velocity component v sub y is shown. The vertical velocity component v sub y is upwards till it reaches the maximum point and then its direction changes to downwards. In part d resultant v of horizontal velocity component v sub x and downward vertical velocity component v sub y is found which makes an angle theta with the horizontal x axis. The direction of resultant velocity v is towards south east.
Figure 3.37 (a) We analyze two-dimensional projectile motion by breaking it into two independent one-dimensional motions along the vertical and horizontal axes. (b) The horizontal motion is simple, because ax=0ax=0 size 12{a rSub { size 8{x} } =0} {} and vxvx size 12{v rSub { size 8{x} } } {} is thus constant. (c) The velocity in the vertical direction begins to decrease as the object rises; at its highest point, the vertical velocity is zero. As the object falls towards the Earth again, the vertical velocity increases again in magnitude but points in the opposite direction to the initial vertical velocity. (d) The x - and y -motions are recombined to give the total velocity at any given point on the trajectory.

Example 3.4 A Fireworks Projectile Explodes High and Away

During a fireworks display, a shell is shot into the air with an initial speed of 70.0 m/s at an angle of 75.0°75.0° above the horizontal, as illustrated in Figure 3.38. The fuse is timed to ignite the shell just as it reaches its highest point above the ground. (a) Calculate the height at which the shell explodes. (b) How much time passed between the launch of the shell and the explosion? (c) What is the horizontal displacement of the shell when it explodes?

Strategy

Because air resistance is negligible for the unexploded shell, the analysis method outlined above can be used. The motion can be broken into horizontal and vertical motions in which ax=0ax=0 size 12{ a rSub { size 8{x} } =0} {} and ay=gay=g size 12{ a rSub { size 8{y} } =-g} {}. We can then define x0x0 size 12{x rSub { size 8{0} } } {} and y0y0 size 12{y rSub { size 8{0} } } {} to be zero and solve for the desired quantities.

Solution for (a)

By “height” we mean the altitude or vertical position yy size 12{y} {} above the starting point. The highest point in any trajectory, called the apex, is reached when vy=0vy=0 size 12{ v rSub { size 8{y} } =0} {}. Since we know the initial and final velocities as well as the initial position, we use the following equation to find yy size 12{y} {}:

vy2=v0y22g(yy0).vy2=v0y22g(yy0). size 12{v rSub { size 8{y} } rSup { size 8{2} } =v rSub { size 8{0y} } rSup { size 8{2} } - 2g \( y - y rSub { size 8{0} } \) "."} {}
3.45
The x y graph shows the trajectory of fireworks shell. The initial velocity of the shell v zero is at angle theta zero equal to seventy five degrees with the horizontal x axis. The fuse is set to explode the shell at the highest point of the trajectory which is at a height h equal to two hundred thirty three meters and at a horizontal distance x equal to one hundred twenty five meters from the origin.
Figure 3.38 The trajectory of a fireworks shell. The fuse is set to explode the shell at the highest point in its trajectory, which is found to be at a height of 233 m and 125 m away horizontally.

Because y0y0 size 12{y rSub { size 8{0} } } {} and vyvy size 12{v rSub { size 8{y} } } {} are both zero, the equation simplifies to

0=v 0y 22gy.0=v 0y 22gy. size 12{0=v rSub { size 8{0y} } rSup { size 8{2} } - 2 ital "gy."} {}
3.46

Solving for yy size 12{y} {} gives

y = v 0y 2 2g . y = v 0y 2 2g . size 12{y= { {v rSub { size 8{0y} } rSup { size 8{2} } } over {2g} } "." } {}
3.47

Now we must find v0yv0y size 12{v rSub { size 8{0y} } } {}, the component of the initial velocity in the y-direction. It is given by v0y=v0sinθv0y=v0sinθ size 12{v rSub { size 8{0y rSup} =v rSub {0 rSup size 12{"sin"θ}} {}, where v0yv0y is the initial velocity of 70.0 m/s, and θ0=75.0°θ0=75.0° size 12{θ rSub { size 8{0} } } {} is the initial angle. Thus,

v0y=v0sinθ0=(70.0 m/s)(sin 75°)=67.6 m/s.v0y=v0sinθ0=(70.0 m/s)(sin 75°)=67.6 m/s. size 12{v rSub { size 8{0y} } =v rSub { size 8{0} } "sin"θ rSub { size 8{0} } = \( "70" "." 0" m/s" \) \( "sin""75" { size 12{ circ } } \) ="67" "." 6" m/s."} {}
3.48

and yy size 12{y} {} is

y = ( 67 .6 m/s ) 2 2 ( 9 . 80 m /s 2 ) , y = ( 67 .6 m/s ) 2 2 ( 9 . 80 m /s 2 ) , size 12{y= { { \( "67" "." 6" m/s" \) rSup { size 8{2} } } over {2 \( 9 "." "80"" m/s" rSup { size 8{2} } \) } } } {}
3.49

so that

y=233 m.y=233 m. size 12{y="233"" m."} {}
3.50

Discussion for (a)

Note that because up is positive, the initial velocity is positive, as is the maximum height, but the acceleration due to gravity is negative. Note also that the maximum height depends only on the vertical component of the initial velocity, so that any projectile with a 67.6 m/s initial vertical component of velocity will reach a maximum height of 233 m (neglecting air resistance). The numbers in this example are reasonable for large fireworks displays, the shells of which do reach such heights before exploding. In practice, air resistance is not completely negligible, and so the initial velocity would have to be somewhat larger than that given to reach the same height.

Solution for (b)

As in many physics problems, there is more than one way to solve for the time to the highest point. In this case, the easiest method is to use y=y0+12(v0y+vy)ty=y0+12(v0y+vy)t size 12{y=y rSub { size 8{0} } + { {1} over {2} } \( v rSub { size 8{0y} } +v rSub { size 8{y} } \) t} {}. Because y0y0 size 12{y rSub { size 8{0} } } {} is zero, this equation reduces to simply

y = 1 2 ( v 0y + v y ) t . y = 1 2 ( v 0y + v y ) t . size 12{y= { {1} over {2} } \( v rSub { size 8{0y} } +v rSub { size 8{y} } \) t} {}
3.51

Note that the final vertical velocity, vyvy size 12{v rSub { size 8{y} } } {}, at the highest point is zero. Thus,

t = 2 y ( v 0y + v y ) = 2 ( 233 m ) ( 67.6 m/s ) = 6.90 s. t = 2 y ( v 0y + v y ) = 2 ( 233 m ) ( 67.6 m/s ) = 6.90 s. alignl { stack { size 12{t= { {2y} over { \( v rSub { size 8{0y} } +v rSub { size 8{y} } \) } } = { {2 times "233"" m"} over { \( "67" "." 6" m/s" \) } } } {} # =6 "." "90"" s" {} } } {}
3.52

Discussion for (b)

This time is also reasonable for large fireworks. When you are able to see the launch of fireworks, you will notice several seconds pass before the shell explodes. (Another way of finding the time is by using y=y0+v0yt12gt2y=y0+v0yt12gt2 size 12{y=y rSub { size 8{0} } +v rSub { size 8{0y} } t - { {1} over {2} } ital "gt" rSup { size 8{2} } } {}, and solving the quadratic equation for tt size 12{t} {}.)

Solution for (c)

Because air resistance is negligible, ax=0ax=0 size 12{a rSub { size 8{x} } =0} {} and the horizontal velocity is constant, as discussed above. The horizontal displacement is horizontal velocity multiplied by time as given by x=x0+vxtx=x0+vxt size 12{x=x rSub { size 8{0} } +v rSub { size 8{x} } t} {}, where x0x0 size 12{x rSub { size 8{0} } } {} is equal to zero:

x=vxt,x=vxt, size 12{x=v rSub { size 8{x} } t ","} {}
3.53

where vxvx size 12{v rSub { size 8{x} } } {} is the x-component of the velocity, which is given by vx=v0cosθ0.vx=v0cosθ0. size 12{v rSub { size 8{x} } =v rSub { size 8{0} } "cos"θ rSub { size 8{0} } "." } {} Now,

vx=v0cosθ0=(70.0 m/s)(cos 75.0°)=18.1 m/s.vx=v0cosθ0=(70.0 m/s)(cos 75.0°)=18.1 m/s. size 12{v rSub { size 8{x} } =v rSub { size 8{0} } "cos"θ rSub { size 12{0} } = \( "70" "." 0" m/s" \) \( "cos""75.0°" \) ="18" "." 1" m/s."} {}
3.54

The time tt size 12{t} {} for both motions is the same, and so xx size 12{t} {} is

x=(18.1 m/s)(6.90 s)=125 m.x=(18.1 m/s)(6.90 s)=125 m. size 12{x= \( "18" "." 1" m/s" \) \( 6 "." "90"" s" \) ="125"" m."} {}
3.55

Discussion for (c)

The horizontal motion is a constant velocity in the absence of air resistance. The horizontal displacement found here could be useful in keeping the fireworks fragments from falling on spectators. Once the shell explodes, air resistance has a major effect, and many fragments will land directly below.

In solving part (a) of the preceding example, the expression we found for yy size 12{y} {} is valid for any projectile motion where air resistance is negligible. Call the maximum height y=hy=h size 12{y=h} {}; then,

h = v 0 y 2 2 g . h = v 0 y 2 2 g . size 12{y= { {v rSub { size 8{0y} } rSup { size 8{2} } } over {2g} } "." } {}
3.56

This equation defines the maximum height of a projectile and depends only on the vertical component of the initial velocity.

Defining a Coordinate System

It is important to set up a coordinate system when analyzing projectile motion. One part of defining the coordinate system is to define an origin for the xx size 12{x} {} and yy size 12{y} {} positions. Often, it is convenient to choose the initial position of the object as the origin such that x0=0x0=0 size 12{x rSub { size 8{0} } =0} {} and y0=0y0=0 size 12{y rSub { size 8{0} } =0} {}. It is also important to define the positive and negative directions in the xx size 12{x} {} and yy size 12{y} {} directions. Typically, we define the positive vertical direction as upwards, and the positive horizontal direction is usually the direction of the object's motion. When this is the case, the vertical acceleration, gg size 12{g} {}, takes a negative value (since it is directed downwards towards the Earth). However, it is occasionally useful to define the coordinates differently. For example, if you are analyzing the motion of a ball thrown downwards from the top of a cliff, it may make sense to define the positive direction downwards since the motion of the ball is solely in the downwards direction. If this is the case, gg size 12{g} {} takes a positive value.

Example 3.5 Calculating Projectile Motion: Hot Rock Projectile

Kilauea in Hawaii is the world's most continuously active volcano. Very active volcanoes characteristically eject red-hot rocks and lava rather than smoke and ash. Suppose a large rock is ejected from the volcano with a speed of 25.0 m/s and at an angle 35.0°35.0° size 12{"35"°} {} above the horizontal, as shown in Figure 3.39. The rock strikes the side of the volcano at an altitude 20.0 m lower than its starting point. (a) Calculate the time it takes the rock to follow this path. (b) What are the magnitude and direction of the rock's velocity at impact?

The trajectory of a rock ejected from a volcano is shown. The initial velocity of rock v zero is equal to twenty five meters per second and it makes an angle of thirty five degrees with the horizontal x axis. The figure shows rock falling down a height of twenty meters below the volcano level. The velocity at this point is v which makes an angle of theta with horizontal x axis. The direction of v is south east.
Figure 3.39 The trajectory of a rock ejected from the Kilauea volcano.

Strategy

Again, resolving this two-dimensional motion into two independent one-dimensional motions will allow us to solve for the desired quantities. The time a projectile is in the air is governed by its vertical motion alone. We will solve for tt size 12{t} {} first. While the rock is rising and falling vertically, the horizontal motion continues at a constant velocity. This example asks for the final velocity. Thus, the vertical and horizontal results will be recombined to obtain vv size 12{v} {} and θvθv size 12{θ rSub { size 8{v} } } {} at the final time tt size 12{t} {} determined in the first part of the example.

Solution for (a)

While the rock is in the air, it rises and then falls to a final position 20.0 m lower than its starting altitude. We can find the time for this by using

y=y0+v0yt12gt2 .y=y0+v0yt12gt2 . size 12{y=y rSub { size 8{0} } +v rSub { size 8{0y} } t - { {1} over {2} } ital "gt" rSup { size 8{2} } "."} {}
3.57

If we take the initial position y0y0 size 12{y rSub { size 8{0} } } {} to be zero, then the final position is y=20.0 m.y=20.0 m. size 12{y= - "20" "." 0" m" "." } {} Now the initial vertical velocity is the vertical component of the initial velocity, found from v0y=v0sinθ0v0y=v0sinθ0 size 12{v rSub { size 8{0y} } =v rSub { size 8{0} } "sin"θ rSub { size 8{0} } } {} = (25.0 m/s25.0 m/s size 12{"25" "." "0 m/s"} {})(sin 35.0°sin 35.0° size 12{"sin 35"°} {}) = 14.3 m/s14.3 m/s size 12{"14" "." "3 m/s"} {}. Substituting known values yields

20.0 m=(14.3 m/s)t4.90 m/s2t2 .20.0 m=(14.3 m/s)t4.90 m/s2t2 . size 12{ - "20" "." 0" m"= \( "14" "." 3" m/s" \) t - left (4 "." "90"" m/s" rSup { size 8{2} } right )t rSup { size 8{2} } "."} {}
3.58

Rearranging terms gives a quadratic equation in tt size 12{t} {}:

4.90 m/s2t214.3 m/st20.0 m=0.4.90 m/s2t214.3 m/st20.0 m=0. size 12{ left (4 "." "90"" m/s" rSup { size 8{2} } right )t rSup { size 8{2} } - left ("14" "." "3 m/s" right )t - left ("20" "." 0" m" right )=0.} {}
3.59

This expression is a quadratic equation of the form at2 + bt + c = 0 at2 + bt + c = 0 size 12{ ital "at" rSup { size 8{2} } + ital "bt"+c=0} {} , where the constants are a=4.90 a=4.90 , b=14.3 b=14.3 , and c=20.0. c=20.0. Its solutions are given by the quadratic formula:

t = b ± b 2 4 ac 2a . t = b ± b 2 4 ac 2a . size 12{t= { { - b +- sqrt {b rSup { size 8{2} } - 4 ital "ac"} } over {"2a"} } "." } {}
3.60

This equation yields two solutions: t=3.96t=3.96 size 12{t=3 "." "96"} {} and t=1.03t=1.03 size 12{t=3 "." "96"} {}. (It is left as an exercise for the reader to verify these solutions.) The time is t=3.96st=3.96s size 12{t=3 "." "96""s"} {} or 1.03s1.03s size 12{-1 "." "03""s"} {}. The negative value of time implies an event before the start of motion, and so we discard it. Thus,

t=3.96 s.t=3.96 s. size 12{t=3 "." "96"" s."} {}
3.61

Discussion for (a)

The time for projectile motion is completely determined by the vertical motion. So any projectile that has an initial vertical velocity of 14.3 m/s and lands 20.0 m below its starting altitude will spend 3.96 s in the air.

Solution for (b)

From the information now in hand, we can find the final horizontal and vertical velocities vxvx size 12{v rSub { size 8{x} } } {} and vyvy size 12{v rSub { size 8{y} } } {} and combine them to find the total velocity vv size 12{v} {} and the angle θ0θ0 size 12{θ rSub { size 8{0} } } {} it makes with the horizontal. Of course, vxvx size 12{v rSub { size 8{x} } } {} is constant so we can solve for it at any horizontal location. In this case, we chose the starting point since we know both the initial velocity and initial angle. Therefore:

vx=v0cosθ0=(25.0 m/s)(cos 35°)=20.5 m/s.vx=v0cosθ0=(25.0 m/s)(cos 35°)=20.5 m/s. size 12{v rSub { size 8{x} } =v rSub { size 8{0} } "cos"θ rSub { size 8{0} } = \( "25" "." 0" m/s" \) \( "cos""35" rSup { size 8{ circ } } \) ="20" "." 5" m/s."} {}
3.62

The final vertical velocity is given by the following equation:

vy=v0ygt,vy=v0ygt, size 12{v rSub { size 8{y} } =v rSub { size 8{0y} } - ital "gt,"} {}
3.63

where v0yv0y size 12{v rSub { size 8{0y} } } {} was found in part (a) to be 14.3 m/s14.3 m/s size 12{"14" "." "3 m/s"} {}. Thus,

v y = 14 . 3 m/s ( 9 . 80 m/s 2 ) ( 3 . 96 s ) v y = 14 . 3 m/s ( 9 . 80 m/s 2 ) ( 3 . 96 s ) size 12{v rSub { size 8{y} } ="14" "." 3" m/s" - \( 9 "." "80"" m/s" rSup { size 8{2} } \) \( 3 "." "96"" s" \) } {}
3.64

so that

vy=24.5 m/s.vy=24.5 m/s. size 12{v rSub { size 8{y} } = - "24" "." 5" m/s."} {}
3.65

To find the magnitude of the final velocity vv size 12{v} {} we combine its perpendicular components, using the following equation:

v=vx2+vy2=(20.5 m/s)2+(24.5 m/s)2,v=vx2+vy2=(20.5 m/s)2+(24.5 m/s)2, size 12{v= sqrt {v rSub { size 8{x} } rSup { size 8{2} } +v rSub { size 8{y} } rSup { size 8{2} } } = sqrt { \( "20" "." 5" m/s" \) rSup { size 8{2} } + \( - "24" "." 5" m/s" \) rSup { size 8{2} } } ","} {}
3.66

which gives

v=31.9 m/s.v=31.9 m/s. size 12{v="31" "." 9" m/s."} {}
3.67

The direction θvθv size 12{θ rSub { size 8{v} } } {} is found from the equation:

θ v = tan 1 ( v y / v x ) θ v = tan 1 ( v y / v x ) size 12{θ rSub { size 8{v} } ="tan" rSup { size 8{ - 1} } \( v rSub { size 8{y} } /v rSub { size 8{x} } \) } {}
3.68

so that

θv=tan1(24.5/20.5)=tan1(1.19).θv=tan1(24.5/20.5)=tan1(1.19). size 12{θ rSub { size 8{v} } ="tan" rSup { size 8{ - 1} } \( - "24" "." 5/"20" "." 5 \) ="tan" rSup { size 8{ - 1} } \( - 1 "." "19" \) "."} {}
3.69

Thus,

θv=50.1°.θv=50.1°. size 12{θ rSub { size 8{v} } = - "50" "." 1 rSup { size 12{ circ } "."} } {}
3.70

Discussion for (b)

The negative angle means that the velocity is 50.50. size 12{"50" "." 1°} {} below the horizontal. This result is consistent with the fact that the final vertical velocity is negative and hence downward—as you would expect because the final altitude is 20.0 m lower than the initial altitude. (See Figure 3.39.)

One of the most important things illustrated by projectile motion is that vertical and horizontal motions are independent of each other. Galileo was the first person to fully comprehend this characteristic. He used it to predict the range of a projectile. On level ground, we define range to be the horizontal distance RR size 12{R} {} traveled by a projectile. Galileo and many others were interested in the range of projectiles primarily for military purposes—such as aiming cannons. However, investigating the range of projectiles can shed light on other interesting phenomena, such as the orbits of satellites around the Earth. Let us consider projectile range further.

Part a of the figure shows three different trajectories of projectiles on level ground. In each case the projectiles makes an angle of forty five degrees with the horizontal axis. The first projectile of initial velocity thirty meters per second travels a horizontal distance of R equal to ninety one point eight meters. The second projectile of initial velocity forty meters per second travels a horizontal distance of R equal to one hundred sixty three meters. The third projectile of initial velocity fifty meters per second travels a horizontal distance of R equal to two hundred fifty five meters.
Figure 3.40 Trajectories of projectiles on level ground. (a) The greater the initial speed v0v0 size 12{v rSub { size 8{0} } } {}, the greater the range for a given initial angle. (b) The effect of initial angle θ0θ0 size 12{θ rSub { size 8{0} } } {} on the range of a projectile with a given initial speed. Note that the range is the same for 15°15° size 12{"15"°} {} and 75°75° size 12{"75°"} {}, although the maximum heights of those paths are different.

How does the initial velocity of a projectile affect its range? Obviously, the greater the initial speed v0v0 size 12{v rSub { size 8{0} } } {}, the greater the range, as shown in Figure 3.40(a). The initial angle θ0θ0 size 12{θ rSub { size 8{0} } } {} also has a dramatic effect on the range, as illustrated in Figure 3.40(b). For a fixed initial speed, such as might be produced by a cannon, the maximum range is obtained with θ0=45°θ0=45° size 12{θ rSub { size 8{0} }  = "45°"} {}. This is true only for conditions neglecting air resistance. If air resistance is considered, the maximum angle is approximately 38°38° size 12{"38°"} {}. Interestingly, for every initial angle except 45°45° size 12{"45°"} {}, there are two angles that give the same range—the sum of those angles is 90°90° size 12{"90°"} {}. The range also depends on the value of the acceleration of gravity gg size 12{g} {}. The lunar astronaut Alan Shepherd was able to drive a golf ball a great distance on the Moon because gravity is weaker there. The range RR size 12{R} {} of a projectile on level ground for which air resistance is negligible is given by

R=v02sin2θ0g,R=v02sin2θ0g, size 12{R= { {v rSub { size 8{0} } rSup { size 8{2} } "sin"2θ rSub { size 8{0} } } over {g} } ","} {}
3.71

where v0v0 size 12{v rSub { size 8{0} } } {} is the initial speed and θ0θ0 size 12{θ rSub { size 8{0} } } {} is the initial angle relative to the horizontal. The proof of this equation is left as an end-of-chapter problem (hints are given), but it does fit the major features of projectile range as described.

When we speak of the range of a projectile on level ground, we assume that RR size 12{R} {} is very small compared with the circumference of the Earth. If, however, the range is large, the Earth curves away below the projectile and acceleration of gravity changes direction along the path. The range is larger than predicted by the range equation given above because the projectile has farther to fall than it would on level ground. (See Figure 3.41.) If the initial speed is great enough, the projectile goes into orbit. This possibility was recognized centuries before it could be accomplished. When an object is in orbit, the Earth curves away from underneath the object at the same rate as it falls. The object thus falls continuously but never hits the surface. These and other aspects of orbital motion, such as the rotation of the Earth, will be covered analytically and in greater depth later in this text.

Once again we see that thinking about one topic, such as the range of a projectile, can lead us to others, such as the Earth orbits. In Addition of Velocities, we will examine the addition of velocities, which is another important aspect of two-dimensional kinematics and will also yield insights beyond the immediate topic.

A figure of the Earth is shown and on top of it a very high tower is placed. A projectile satellite is launched from this very high tower with initial velocity of v zero in the horizontal direction. Several trajectories are shown with increasing range. A circular trajectory is shown indicating the satellite achieved its orbit and it is revolving around the Earth.
Figure 3.41 Projectile to satellite. In each case shown here, a projectile is launched from a very high tower to avoid air resistance. With increasing initial speed, the range increases and becomes longer than it would be on level ground because the Earth curves away underneath its path. With a large enough initial speed, orbit is achieved.

PhET Explorations: Projectile Motion

Blast a Buick out of a cannon! Learn about projectile motion by firing various objects. Set the angle, initial speed, and mass. Add air resistance. Make a game out of this simulation by trying to hit a target.

Figure 3.42 Projectile Motion
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