2. Two Dimensional Motion Two components: Horizontal (x-axis) & Vertical (y-axis)

3. Ex:1) projectile fired from a gun 2) car turning around a corner 3) child swinging on a playground

4. Independence of Motions Vector Components taken perpendicular to each other are independent. ***horizontal & vertical motions are independent***

5. If you fired a gun horizontally and dropped a similar bullet from the same height, at the same time as the other bullet left the barrel, which bullet would hit the ground first?

6. Ex- a bullet fired from a gun falls at the same rate & hits the ground at the same time as a similar bullet simply dropped from the muzzle

8. Projectiles Objects that are thrown or launched into the air and are subject to gravity.

9. Trajectory A curved path, called a parabola, that a projectile follows.

10. Air resistance slows down projectile, as it collides with air molecules. Thereby diminishing the trajectory.

11. We are going to neglect the effects of air resistance

12. Projectiles fired horizontally.

15. Ex: Let’s analyze a bullet fired from a gun from a height of 80 meters with an initial velocity of 200 m/s.

16. Horizontal: 1 st the gun exerts a force on the bullet, giving it motion - let v i,x = 200 m/s

17. Are there any horizontal forces to change the bullets speed once it is shot? NO!

18. remains constant

19. Vertical: = 0 m/s bullet fired horizontally What vertical forces act on the bullet once it is shot? Gravity = - 10 m/s 2

20. What happens to the bullet’s velocity as it falls? Increases negatively with time due to gravity

21. Recall frame of reference: To locate bullet at any time,  t, after it is shot…

22. To figure out the displacement Horizontal (  x): there is no acceleration once the bullet leaves the barrel.

24. Once the projectile is launched, it no longer undergoes acceleration. Therefore, the horizontal velocity (v x ) remains constant.

25. v x = v i,x = constant

26. Vertical (  y): when the bullet leaves the barrel there is no vertical velocity initially, but gravity takes right a hold of it.

27. acceleration is gravity

28. v f,y = g  t v f,y 2 = 2 g  y

29. time  x = v xi  t  y = ½ g  t 2 y = height+  y (s) (m) (m) (m) 0 1 2 3 4 0 200 400 600 800 0 - 5 - 20 - 45 - 80 80 75 60 35 0

30. Projectiles Fired Horizontally

31. Projectiles fired horizontally at different speeds

32. Ex : Joe kicks a 0.75 kg ball off a 50 m cliff. If he gives the ball only an initial horizontal speed of 15 m/s. A) How long does it take to hit the ground? B) how far does it travel in the horizontal direction?

33. G: m =0.75 kg,  y= -50 m v i,x =15 m/s, g= -10 m/s 2 U:  t = ?,  x =?

34. G: m =0.75 kg,v i,x =15m/s, g= - 10 m/s 2,  y= - 50 m U:  t = ?,  x =? E: A)

35. S:

37. E: B)  x = v i,x  t S:  x = (15 m/s) x (3.16 s) S:  x = 47.4 m

38. Ex 2: A golf ball is hit off a cliff at a speed of 80 m/s horizontally and travels 471 m. A) Calculate the height of the cliff.

39. G: g = -10 m/s 2,  x = 471 m, v i,x = 80 m/s U:  y = ?

40. G: g = - 10 m/s 2,  x = 471 m, v i,x =80 m/s U:  y=? E:

41. G: g = -10 m/s 2,  x = 471 m v i,x = 80 m/s U:  y = ? E: Need to find  t

42.  t =  x / v i,x  t = 471 m / 80 m/s  t = 5.89 sec

43. S:  y = ½ (-10)(5.89) 2 S:  y = -173.5 m

44. B) How fast will the ball hit the ground? Since the ball is moving in both the x and y directions we need to combine both velocities.

45. Find v f,y v f,y = g  t v f,y =(-10 m/s 2 )(5.89 s) v f,y = - 58.9 m/s

46. E:

47. S:

48. E: S:

49. Projectile Fired at an Angle When an object is thrown up or at an angle, the initial y- component velocity decreases due to gravity.

50. At its peak the ‘y’ velocity is zero, acceleration is still –10 m/s 2. The object begins to fall due to gravity, increasing its speed, but NEGATIVE.

51. When it reaches the point where it was launched from, its velocity is the same as the initial y component, but negative.

54. Ex 3: Consider a cannon firing at a 30 o angle with an initial speed of 39.2 m/s.

55. Remember to consider independent components separately!

56. Horizontal Formulas :

57. From vectors, how did we find the x component of vector at an angle?

58. Horizontal Formulas :

60. Remember horizontal motion is constant.

61. Recall, no forces change horizontal motion …therefore

63. Vertical:

64. Formulas : From vectors, how do we find the y component of a vector at an angle?

65. Vertical: Formulas :

66. Vertical: Formulas :

67. v f,y = v i (sin  )+g  t v f,y 2 = v i 2 (sin  ) 2 + 2g  y

68. The cannon ball decelerates to a maximum height where and then accelerates to the ground.

70. Horizontal Distance Traveled by a Projectile Range- horizontal distance traveled by a projectile

71. -Maximum range achieved with a firing angle of 45 degrees.

72. Actual Motion of a Projectile Simplifying assumptions: 1)Air resistance can be neglected 2)Earth can be considered flat for small ranges

73. 3) For short ranges, gravity acts in the same direction constantly 4) Projectiles remain in the same vertical plane throughout flight

74. Effect of these assumptions: projectile turns downward from it’s ideal path Thus, it has a shorter range

75. Monkey and the Hunter

76. http://www.phys.unsw.edu.au/~jw/d emo/projectiles.html http://www.phy.ntnu.edu.tw/ntnujava/index. php?topic=144.0http://www.phy.ntnu.edu.tw/ntnujava/index. php?topic=144.0

77. http://www.phys.unsw.edu.au/~jw/d emo/projectiles.html http://www.phy.ntnu.edu.tw/ntnujava/index. php?topic=144.0http://www.phy.ntnu.edu.tw/ntnujava/index. php?topic=144.0

78. When a projectile lands or is caught at the same height as it was launched: 1.The vertical displacement is zero.  y= 0 m 2.The final ‘y’ velocity is the negative of the initial ‘y’ velocity. v f,y = -v i,y

79. If you are given the initial velocity @ an angle always find the x and y components of the velocity.

80. Ex 4: A baseball is hit 45 m/s at 45 o. A) How long is the ball in the air? B) How far does the ball travel horizontally?

81. G: v i = 45 m/s,  =45 o g = -10 m/s 2 U:  t = ?,  x =? Find the ‘x’&‘y’ velocity components first.

82. Since the angle is 45 o both x and y components are the same. (forms an isosceles triangle)

83. E: v i,x = v i cos  S: v i,x = 45cos(45) S: v i,x = 31.8 m/s v i,x = v i,y = 31.8 m/s

84. E:

86. Remember the final ‘y’ velocity is the negative of the initial ‘y’ velocity.

87. S:

88. S:  t = 6.36 sec

89. E:

90. E: S:

91. E: S:

92. EX 5: In a scene from an action movie a stunt man must leap from one building to another that is 4 m apart. After a running start the actor leaps from the roof at an angle of 15 o with a speed of 5 m/s. Will he make the it to the other roof, which is 2.5 m shorter than the one he jumped from?

93. G:  y = - 2.5 m, g = -10m/s 2, v i = 5 m/s,  = 15 o U:

94. When you : Range (  x) Height (  y) Time (  t) Final Velocity Increase Mass Increase Velocity Increase Angle Add Air Resistance

95. When you : Range (  x) Height (  y) Time (  t) Final Velocity Increase Mass No Change Increase Velocity Increase Angle Add Air Resistance

96. When you : Range (  x) Height (  y) Time (  t) Final Velocity Increase Mass No Change Increase Velocity Increase Angle Add Air Resistance

97. When you : Range (  x) Height (  y) Time (  t) Final Velocity Increase Mass No Change Increase Velocity Increase Angle Add Air Resistance

98. When you : Range (  x) Height (  y) Time (  t) Final Velocity Increase Mass No Change Increase Velocity Increase Angle Add Air Resistance

99. When you : Range (  x) Height (  y) Time (  t) Final Velocity Increase Mass No Change Increase Velocity Increases Increase Angle Add Air Resistance

100. When you : Range (  x) Height (  y) Time (  t) Final Velocity Increase Mass No Change Increase Velocity Increases Increase Angle Add Air Resistance

101. When you : Range (  x) Height (  y) Time (  t) Final Velocity Increase Mass No Change Increase Velocity Increases Increase Angle Add Air Resistance

102. When you : Range (  x) Height (  y) Time (  t) Final Velocity Increase Mass No Change Increase Velocity Increases Note 1 Increase Angle Add Air Resistance

103. When you : Range (  x) Height (  y) Time (  t) Final Velocity Increase Mass No Change Increase Velocity Increases Note 1 Increase Angle Note 2 Add Air Resistance

104. When you : Range (  x) Height (  y) Time (  t) Final Velocity Increase Mass No Change Increase Velocity Increases Note 1 Increase Angle Note 2 Increases Note 3 Add Air Resistance

105. When you : Range (  x) Height (  y) Time (  t) Final Velocity Increase Mass No Change Increase Velocity Increases Note 1 Increase Angle Note 2 Increases Note 3 Increases Note 4 Add Air Resistance

106. When you : Range (  x) Height (  y) Time (  t) Final Velocity Increase Mass No Change Increase Velocity Increases Note 1 Increase Angle Note 2 Increases Note 3 Increases Note 4 Stays the Same Add Air Resistance

107. When you : Range (  x) Height (  y) Time (  t) Final Velocity Increase Mass No Change Increase Velocity Increases Note 1 Increase Angle Note 2 Increases Note 3 Increases Note 4 Stays the Same Add Air Resistance Decreases

108. When you : Range (  x) Height (  y) Time (  t) Final Velocity Increase Mass No Change Increase Velocity Increases Note 1 Increase Angle Note 2 Increases Note 3 Increases Note 4 Stays the Same Add Air Resistance Decreases

109. When you : Range (  x) Height (  y) Time (  t) Final Velocity Increase Mass No Change Increase Velocity Increases Note 1 Increase Angle Note 2 Increases Note 3 Increases Note 4 Stays the Same Add Air Resistance Decreases

110. When you : Range (  x) Height (  y) Time (  t) Final Velocity Increase Mass No Change Increase Velocity Increases Note 1 Increase Angle Note 2 Increases Note 3 Increases Note 4 Stays the Same Add Air Resistance Decreases

111. Note 1 As initial velocity changes, the final velocity follows. (It’s the same)

112. Note 2 Increasing the launch angle up to 45 degrees increases the range.

113. After 45 degrees, the range decreases. 45 degrees gives you the maximum range.

114. Complimentary angles (i.e. 15 & 75) give you the same range.

115. 90 degrees gives you the maximum height. Note 3

116. Note 4 90 degrees gives you the maximum ‘hang’ time.