1. Motion in Two Dimensions Chapter 6

2. Projectile Motion Recognize that the vertical and horizontal motions of a projectile are independent. Relate the height, time in the air, and initial vertical velocity of a projectile using its vertical motion, and then determine the range using the horizontal motion. Explain how the trajectory of a projectile depends upon the frame of reference from which it is observed. In this section you will: Section 6.1

3. Projectile Motion If you observed the movement of a golf ball being hit from a tee, a frog hopping, or a free throw being shot with a basketball, you would notice that all of these objects move through the air along similar paths, as do baseballs, arrows, and bullets. Each path is a curve that moves upward for a distance, and then, after a time, turns and moves downward for some distance. You may be familiar with this curve, called a parabola Projectile Motion Section 6.1

4. Projectile Motion An object shot through the air is called a projectile. Projectile Motion Section 6.1 A projectile can be a football, a bullet, or a drop of water. The force of gravity is what causes the object to curve downward in a parabolic flight path. Its path through space is called its trajectory.

5. Independence of Motion in Two Dimensions Section 6.1 Projectile Motion

6. When a projectile is launched at an angle, the initial velocity has a vertical component as well as a horizontal component. If the object is launched upward, like a ball tossed straight up in the air, it rises with slowing speed, reaches the top of its path, and descends with increasing speed. Projectiles Launched at an Angle Section 6.1

7. Projectile Motion At each point in the vertical direction, the velocity of the object as it is moving upward has the same magnitude as when it is moving downward. The only difference is that the directions of the two velocities are opposite. Projectiles Launched at an Angle Section 6.1

8. Projectile Motion The adjoining figure defines two quantities associated with a trajectory. One is the maximum height, which is the height of the projectile when the vertical velocity is zero. Projectiles Launched at an Angle Section 6.1

9. Projectile Motion The other quantity depicted is the range, R, which is the horizontal distance that the projectile travels. Time is how much time the projectile is in the air. Flight time often is called hang time. Projectiles Launched at an Angle Section 6.1

10. Choosing a coordinate system, such as the one shown below, is similar to laying a grid. You have to choose where to put the center of the grid (the origin) and establish the directions in which the axes point. You then have X’s and Y’s for components. Vectors Components of Vectors Section 5.1

11. Vectors When the motion is on the surface of Earth, it is often convenient to have the x-axis point east and the y-axis point north. When the motion involves an object moving through the air, the positive x-axis is often chosen to be horizontal to the right and the positive y-axis vertical (upward). If the motion is on a hill, it’s convenient to place the positive x- axis in the direction of the motion (along the ramp) and the y- axis perpendicular to the x-axis (normal) Components of Vectors Section 5.1

12. Vectors Component Vectors Section 5.1

13. Two or more vectors (A, B, C, etc.) may be added by first resolving each vector into its x- and y-components. The x-components are added to form the x-component of the resultant: R x = A x + B x + C x. Similarly, the y-components are added to form the y-component of the resultant: R y = A y + B y + C y. Vectors Algebraic Addition of Vectors Section 5.1

14. Vectors Because R x and R y are at a right angle (90°), the magnitude of the resultant vector can be calculated using the Pythagorean theorem, R 2 = R x 2 + R y 2. Algebraic Addition of Vectors Section 5.1

15. Vectors To find the angle or direction of the resultant, recall that the tangent of the angle that the vector makes with the x-axis is give by the following. Angle of the resultant vector Algebraic Addition of Vectors Section 5.1 The angle of the resultant vector is equal to the inverse tangent of the quotient of the y-component divided by the x- component of the resultant vector.

16. Vectors You can find the angle by using the tan −1 key on your calculator. Note that when tan θ > 0, most calculators give the angle between 0° and 90°, and when tan θ < 0, the angle is reported to be between 0° and −90°. You will use these techniques to resolve vectors into their components throughout your study of physics. Resolving vectors into components allows you to analyze complex systems of vectors without using graphical methods. Algebraic Addition of Vectors Section 5.1

17. Projectile Motion Angle - SOH, CAH, TOA Magnitude – Hypotenuse C 2 = A 2 + B 2 Or  R 2 = R x 2 + R y 2 Magnitude – Component R x = Rcosθ R y = Rsinθ Vector Components – Trig Review (or new material) Your book – Chapter 5.1 Polar CS vs. Cartesian CS Section 5.1

18. Vectors Vector Direction – Cardinal Directions Section 5.1 N S E W N of E E of N S of W W of S N of W W of N S of E E of S

19. Section Check Jeff moved 3 m due north, and then 4 m due west to his friends house. What is the displacement of Jeff? Question 1 Section 5.1 A. 3 + 4 m B. 4 – 3 m C. 3 2 + 4 2 m D. 5 m

20. Section Check Answer: D Answer 1 Section 5.1 Reason: When two vectors are at right angles to each other as in this case, we can use the Pythagorean theorem of vector addition to find the magnitude of resultant, R.

21. Section Check Answer: D Answer 1 Section 5.1 Reason: Pythagorean theorem of vector addition states If vector A is at right angle to vector B then the sum of squares of magnitudes is equal to square of magnitude of resultant vector. That is, R 2 = A 2 + B 2  R 2 = (3 m) 2 + (4 m) 2 = 5 m

22. Calculate the resultant of three vectors A, B, C as shown in the figure. (A x = B x = C x = A y = C y = 1 units and B y = 2 units) Question 2 Section 5.1 Section Check A. 3 + 4 units B. 3 2 + 4 2 units C. 4 2 – 3 2 units D.

23. Section Check Answer: D Answer 2 Section 5.1 Reason: Add the x-components to form, R x = A x + B x + C x. Add the y-components to form, R y = A y + B y + C y.

24. Section Check Answer: D Answer 2 Section 5.1 Reason: Since R x and R y are perpendicular to each other we can apply Pythagoras theorem of vector addition: R 2 = R x 2 + R y 2

25. Section Check If a vector B is resolved into two components B x and B y and if  is angle that vector B makes with the positive direction of x-axis, Using which of the following formula can you calculate the components of vector B? Question 3 Section 5.1 A. B x = B sin θ, B y = B cos θ B. B x = B cos θ, B y = B sin θ C. D.

26. Section Check Answer: B Answer 3 Section 5.1 Reason: The components of vector ‘B’ are calculated using the equation stated below.

27. Section Check Answer: B Answer 3 Section 5.1 Reason:

28. Projectile Motion The Flight of a Ball A ball is launched at 4.5 m/s at 66° above the horizontal. What are the maximum height and flight time of the ball? Section 6.1

29. The Flight of a Ball Establish a coordinate system with the initial position of the ball at the origin. Projectile Motion Section 6.1

30. The Flight of a Ball Show the positions of the ball at the beginning, at the maximum height, and at the end of the flight. Projectile Motion Section 6.1

31. The Flight of a Ball Draw a motion diagram showing v, a, and F net. Projectile Motion Section 6.1 Known: y i = 0.0 m θ i = 66° v i = 4.5 m/s a y = −g Unknown: y max = ? t = ?

32. The Flight of a Ball Find the y-component of v i. Projectile Motion Section 6.1

33. The Flight of a Ball Substitute v i = 4.5 m/s, θ i = 66° Projectile Motion Section 6.1

34. The Flight of a Ball Find an expression for time. Projectile Motion Section 6.1

35. The Flight of a Ball Substitute a y = −g Projectile Motion Section 6.1

36. The Flight of a Ball Solve for t. Section 6.1 Projectile Motion

37. The Flight of a Ball Solve for the maximum height. Section 6.1 Projectile Motion

38. The Flight of a Ball Section 6.1 Projectile Motion

39. The Flight of a Ball Substitute y i = 0.0 m, v yi = 4.1 m/s, v y = 0.0 m/s at y max, g = 9.80 m/s 2 Projectile Motion Section 6.1

40. The Flight of a Ball Solve for the time to return to the launching height. Proof – Quadratic Formula Projectile Motion Section 6.1

41. The Flight of a Ball Substitute y f = 0.0 m, y i = 0.0 m, a = −g Projectile Motion Section 6.1

42. The Flight of a Ball Use the quadratic formula to solve for t. Projectile Motion Section 6.1

43. The Flight of a Ball 0 is the time the ball left the launch, so use this solution. Projectile Motion Section 6.1

44. The Flight of a Ball Substitute v yi = 4.1 m/s, g = 9.80 m/s 2 Projectile Motion Section 6.1

45. Projectile Motion The path of the projectile, or its trajectory, depends upon who is viewing it. Suppose you toss a ball up and catch it while riding in a bus. To you, the ball would seem to go straight up and straight down. But an observer on the sidewalk would see the ball leave your hand, rise up, and return to your hand, but because the bus would be moving, your hand also would be moving. The bus, your hand, and the ball would all have the same horizontal velocity. Trajectories Depend upon the Viewer Section 6.1

46. Section Check A boy standing on a balcony drops one ball and throws another with an initial horizontal velocity of 3 m/s. Which of the following statements about the horizontal and vertical motions of the balls is correct? (Neglect air resistance.) Question 1 Section 6.1 A. The balls fall with a constant vertical velocity and a constant horizontal acceleration. B. The balls fall with a constant vertical velocity as well as a constant horizontal velocity. C. The balls fall with a constant vertical acceleration and a constant horizontal velocity. D. The balls fall with a constant vertical acceleration and an increasing horizontal velocity.

47. Section Check Answer: C Answer 1 Section 6.1 Reason: The vertical and horizontal motions of a projectile are independent. The only force acting on the two balls is force due to gravity. Because it acts in the vertical direction, the balls accelerate in the vertical direction. The horizontal velocity remains constant throughout the flight of the balls.

48. Section Check Which of the following conditions is met when a projectile reaches its maximum height? Question 2 Section 6.1 A. Vertical component of the velocity is zero. B. Vertical component of the velocity is maximum. C. Horizontal component of the velocity is maximum. D. Acceleration in the vertical direction is zero.

49. Section Check Answer: A Answer 2 Section 6.1 Reason: The maximum height is the height at which the object stops its upward motion and starts falling down, i.e. when the vertical component of the velocity becomes zero.

50. Section Check Suppose you toss a ball up and catch it while riding in a bus. Why does the ball fall in your hands rather than falling at the place where you tossed it? Question 3 Section 6.1

51. Section Check No air resistance – nothing to slow it down. Trajectory depends on the frame of reference. Also inertia (Newton’s Laws) Answer 3 Section 6.1

52. Circular Motion This section will not be covered at this time. Section 6.2

53. Relative Velocity Analyze situations in which the coordinate system is moving. Solve relative-velocity problems. In this section you will: Section 6.3

54. Relative Velocity You are in a school bus that is traveling at a velocity of 8 m/s. You walk with a velocity of 3 m/s. If the bus is traveling at 8 m/s, this means that the velocity of the bus is 8 m/s, as measured by your friend in a coordinate system fixed to the road. Relative Velocity Section 6.3

55. Relative Velocity When you are standing still, your velocity relative to the road is also 8 m/s, but your velocity relative to the bus is zero. A vector representation of this problem is shown in the figure. Relative Velocity Section 6.3

56. Relative Velocity When a coordinate system is moving, two velocities are added if both motions are in the same direction. One is subtracted from the other if the motions are in opposite directions. You will find that your velocity relative to the street is 11 m/s, the sum of 8 m/s and 3 m/s. From the frame of reference to someone in the bus you are moving 3 m/s. Relative Velocity Section 6.3

57. Relative Velocity Two velocities are in opposite directions, the resultant velocity is 5 m/s–the difference between 8 m/s and 3 m/s. Simple addition or subtraction can be used to determine the relative velocity. Relative Velocity Section 6.3

58. Relative Velocity Mathematically, relative velocity is represented as v y/b + v b/r = v y/r. Y=you, b=bus, and r=road The more general form of this equation is: Relative Velocity Section 6.3

59. The method for adding relative velocities also applies to motion in two dimensions. Relative Velocity Section 6.3 Airline pilots must take into account the plane’s speed relative to the air, and their direction of flight relative to the air. They also must consider the velocity of the wind at the altitude they are flying relative to the ground.

60. Relative Velocity Relative Velocity of a Marble Ana and Sandra are riding on a ferry boat that is traveling east at a speed of 4.0 m/s. Sandra rolls a marble with a velocity of 0.75 m/s north, straight across the deck of the boat to Ana. What is the velocity of the marble relative to the water? Section 6.3

61. Relative Velocity of a Marble Establish a coordinate system. Relative Velocity Section 6.3

62. Relative Velocity of a Marble Draw vectors to represent the velocities of the boat relative to the water and the marble relative to the boat. Relative Velocity Section 6.3

63. Relative Velocity of a Marble Identify known and unknown variables. Relative Velocity Section 6.3 Known: v b/w = 4.0 m/s v m/b = 0.75 m/s Unknown: v m/w = ?

64. Relative Velocity of a Marble Because the two velocities are at right angles, use the Pythagorean theorem. Relative Velocity Section 6.3

65. Relative Velocity of a Marble Substitute v b/w = 4.0 m/s, v m/b = 0.75 m/s Relative Velocity Section 6.3

66. Relative Velocity of a Marble Find the angle of the marble’s motion. Relative Velocity Section 6.3

67. Relative Velocity of a Marble Substitute v b/w = 4.0 m/s, v m/b = 0.75 m/s Relative Velocity Section 6.3 = 11° north of east The marble is traveling 4.1 m/s at 11° north of east.

68. Relative Velocity Are the units correct? Dimensional analysis verifies that the velocity is in m/s. Do the signs make sense? The signs should all be positive. Are the magnitudes realistic? The resulting velocity is of the same order of magnitude as the velocities given in the problem. Relative Velocity of a Marble Section 6.3

69. Relative Velocity You can add relative velocities even if they are at arbitrary angles by using the graphical methods. Analyzing a two-dimensional relative- velocity situation is drawing the proper triangle to represent the velocities. Once you have this triangle, you simply apply your knowledge of vector addition. If two velocities are perpendicular to each other, you can find the third by applying the Pythagorean theorem. If the situation has no right angles, you will need to use one or both of the laws of sines and cosines. You will rarely have to do this in HS physics. Relative Velocity Section 6.3

70. Section Check Steven is walking on the roof of a bus with a velocity of 2 m/s toward the rear end of the bus. The bus is moving with a velocity of 10 m/s. What is the velocity of Steven with respect to Anudja sitting inside the bus and Mark standing on the street? Question 1 Section 6.3 A. Velocity of Steven with respect to Anudja is 2 m/s and with respect to Mark is 12 m/s. B. Velocity of Steven with respect to Anudja is 2 m/s and with respect to Mark is 8 m/s. C. Velocity of Steven with respect to Anudja is 10 m/s and with respect to Mark is 12 m/s. D. Velocity of Steven with respect to Anudja is 10 m/s and with respect to Mark is 8 m/s.

71. Section Check Answer: B Answer 1 Section 6.3 Reason: The velocity of Steven with respect to Anudja is 2 m/s since Steven is moving with a velocity of 2 m/s with respect to the bus, and Anudja is at rest with respect to the bus. The velocity of Steven with respect to Mark can be understood with the help of the following vector representation.

72. Section Check Which of the following formulas is the general form of relative velocity of objects a, b, and c? Question 2 Section 6.3 A.V a/b + V a/c = V b/c B.V a/b  V b/c = V a/c C.V a/b + V b/c = V a/c D.V a/b  V a/c = V b/c

73. Section Check Answer: C Answer 2 Section 6.3 Reason: Relative velocity law is V a/b + V b/c = V a/c. The relative velocity of object a to object c is the vector sum of object a’s velocity relative to object b and object b’s velocity relative to object c.

74. Section Check An airplane flies due south at 100 km/hr relative to the air. Wind is blowing at 20 km/hr to the west relative to the ground. What is the plane’s speed with respect to the ground? Question 3 Section 6.3 A. (100 + 20) km/hr B. (100 − 20) km/hr C. D.

75. Section Check Answer: C Answer 3 Section 6.3 Reason: Since the two velocities are at right angles, we can apply Pythagoras theorem of addition law. By using relative velocity law, we can write:

76. End of Chapter 6 Chapter Motion in Two Dimensions