1. Copyright © by Holt, Rinehart and Winston. All rights reserved. Scalars and Vectors

2. Copyright © by Holt, Rinehart and Winston. All rights reserved. Concept Check - Vectors If two vectors are given such that A + B = 0, what can you say about the magnitude and direction of vectors A and B ? 1) same magnitude, but can be in any direction 2) same magnitude, but must be in the same direction 3) different magnitudes, but must be in the same direction 4) same magnitude, but must be in opposite directions 5) different magnitudes, but must be in opposite directions

3. Copyright © by Holt, Rinehart and Winston. All rights reserved. Concept Check - Vectors If two vectors are given such that A + B = 0, what can you say about the magnitude and direction of vectors A and B ? 1) same magnitude, but can be in any direction 2) same magnitude, but must be in the same direction 3) different magnitudes, but must be in the same direction 4) same magnitude, but must be in opposite directions 5) different magnitudes, but must be in opposite directions The magnitudes must be the same, but one vector must be pointing in the opposite direction of the other, in order for the sum to come out to zero. You can prove this with the tip-to-tail method.

4. Copyright © by Holt, Rinehart and Winston. All rights reserved. Concept Check – Vector Addition You are adding vectors of length 20 and 40 units. What is the only possible resultant magnitude that you can obtain out of the following choices? 1) 0 2) 18 3) 37 4) 64 5) 100

5. Copyright © by Holt, Rinehart and Winston. All rights reserved. Concept Check – Vector Addition You are adding vectors of length 20 and 40 units. What is the only possible resultant magnitude that you can obtain out of the following choices? 1) 0 2) 18 3) 37 4) 64 5) 100 minimum opposite20 unitsmaximum aligned 60 units The minimum resultant occurs when the vectors are opposite, giving 20 units. The maximum resultant occurs when the vectors are aligned, giving 60 units. Anything in between is also possible, for angles between 0° and 180°.

6. Copyright © by Holt, Rinehart and Winston. All rights reserved. Compass Rose

7. Copyright © by Holt, Rinehart and Winston. All rights reserved. Adding Vectors Mathematically The Pythagorean Theorem The Tangent Function

8. Copyright © by Holt, Rinehart and Winston. All rights reserved.

9.  v r = 410 km/h @ 21 ° E of N (or 69 ° N of E)

10. Copyright © by Holt, Rinehart and Winston. All rights reserved. Sample Problem Finding Resultant Magnitude and Direction An archaeologist climbs the Great Pyramid in Giza, Egypt. The pyramid’s height is 136 m and its width is 2.30  10 2 m. What is the magnitude and the direction of the displacement of the archaeologist after she has climbed from the bottom of the pyramid to the top?

11. Copyright © by Holt, Rinehart and Winston. All rights reserved. Sample Problem, continued 1. Define Given:  y = 136 m  x = 1/2(width) = 115 m Unknown: d = ?  = ? Diagram: Choose the archaeologist’s starting position as the origin of the coordinate system, as shown above.

12. Copyright © by Holt, Rinehart and Winston. All rights reserved. Sample Problem, continued Rearrange the equations to isolate the unknowns: 2. Plan Choose an equation or situation: The Pythagorean theorem can be used to find the magnitude of the archaeologist’s displacement. The direction of the displacement can be found by using the inverse tangent function.

13. Copyright © by Holt, Rinehart and Winston. All rights reserved. Sample Problem, continued 3. Calculate Evaluate Because d is the hypotenuse, the archaeologist’s displacement should be less than the sum of the height and half of the width. The angle is expected to be more than 45  because the height is greater than half of the width.

14. Copyright © by Holt, Rinehart and Winston. All rights reserved. Resolving Vectors into Components, cont.

15. Copyright © by Holt, Rinehart and Winston. All rights reserved. Sample

16. Copyright © by Holt, Rinehart and Winston. All rights reserved. Problems – 15

17. Copyright © by Holt, Rinehart and Winston. All rights reserved. Adding Vectors That Are Not Perpendicular x1x1 x2x2  x 1,x  x 1,y  x 2,x  x 2,y x1x1 x2x2

18. Copyright © by Holt, Rinehart and Winston. All rights reserved. Adding Non-Perpendicular Vectors 1.Resolve all vectors into x- and y-components. 2.Add x-components, taking into account direction of each. 3.Add y-components in the same way. 4.Vectors resulting from Steps 2 and 3 are ┴ and make up 2 sides of a right triangle whose hypotenuse is the resultant sum of the original vectors. Add them as you would any two ┴ vectors, using Pythagoras and tan –1. x1x1 x2x2  x 1,x  x 1,y  x 2,x  x 2,y

19. Copyright © by Holt, Rinehart and Winston. All rights reserved.  x 1 = 65 m @ 75 o (15° North of East)  x 2 = 48 m @ 27 o (27° East of North)  x r = ? xrxr  x r = 110 m 35° scale: 1 cm = 10 m x1x1 x2x2 @ 35 o North of East 90 o 0o0o 0o0o Adding Vectors Graphically 90 o 270 o 0o0o 180 o A bird dog on the trail of a grouse walks 65 m along a heading of 75°, then turns to walk 48 m along a heading of 27°. Use graphical techniques to find the resultant displacement of the dog.

20. Copyright © by Holt, Rinehart and Winston. All rights reserved.  x 1 = 65 m 15 o  x 2 = 48 m 63 o  x 1,x  x 1,y  x 2,y  x 2,x  x 1 = 65 m @ 75 o (15° North of East)  x 2 = 48 m @ 27 o (27° East of North)  x r = ? 90 o 270 o 0o0o 180 o Adding Vectors Algebraically

21. Copyright © by Holt, Rinehart and Winston. All rights reserved. (continued)  x 1,x x1x1 x2x2 xrxr  x 2,x  y 1,y  y 2,y  yy xx

22. Copyright © by Holt, Rinehart and Winston. All rights reserved. Sample Problem Adding Vectors Algebraically A hiker walks 27.0 km from her base camp at 35° south of east. The next day, she walks 41.0 km in a direction 65° north of east and discovers a forest ranger’s tower. Find the magnitude and direction of her resultant displacement

23. Copyright © by Holt, Rinehart and Winston. All rights reserved. 1. Select a coordinate system. Then sketch and label each vector. Given: d 1 = 27.0 km  1 = –35° d 2 = 41.0 km  2 = 65° Tip:  1 is negative, because clockwise movement from the positive x-axis is negative by convention. Unknown: d = ?  = ? Sample Problem, continued

24. Copyright © by Holt, Rinehart and Winston. All rights reserved. 2. Find the x and y components of all vectors. Make a separate sketch of the displacements for each day. Use the cosine and sine functions to find the components. Sample Problem, continued

25. Copyright © by Holt, Rinehart and Winston. All rights reserved. 3. Find the x and y components of the total displacement. 4. Use the Pythagorean theorem to find the magnitude of the resultant vector. Sample Problem, continued

26. Copyright © by Holt, Rinehart and Winston. All rights reserved. 5. Use a suitable trigonometric function to find the angle. Sample Problem, continued

27. Copyright © by Holt, Rinehart and Winston. All rights reserved. Problems - 20 30 ° 60 ° d2d2 d1d1 d 1,y d 1,x d 2,y d 2,x 30 ° 60 ° d 1,x + d 2,x d 1,y + d 2,y drdr 

28. Copyright © by Holt, Rinehart and Winston. All rights reserved. Projectiles © 1996 Bill Amend by Universal Press Syndicate

29. Copyright © by Holt, Rinehart and Winston. All rights reserved. Concept Check – Projectile Motion A small cart is rolling at constant velocity on a flat track. It fires a ball straight up into the air as it moves. After it is fired, what happens to the ball? 1)it depends on how fast the cart is moving 2) it falls behind the cart 3) it falls in front of the cart 4) it falls right back into the cart 5) it remains at rest

30. Copyright © by Holt, Rinehart and Winston. All rights reserved. Concept Check – Projectile Motion A small cart is rolling at constant velocity on a flat track. It fires a ball straight up into the air as it moves. After it is fired, what happens to the ball? 1)it depends on how fast the cart is moving 2) it falls behind the cart 3) it falls in front of the cart 4) it falls right back into the cart 5) it remains at rest when viewed from train when viewed from ground vertical same horizontal velocity In the frame of reference of the cart, the ball only has a vertical component of velocity. So it goes up and comes back down. To a ground observer, both the cart and the ball have the same horizontal velocity, so the ball still returns into the cart.

31. Copyright © by Holt, Rinehart and Winston. All rights reserved. Concept Check – Horizontal Projectiles You drop a package from a plane flying at constant speed in a straight line. Without air resistance, the package will: 1) quickly lag behind the plane while falling 2) remain vertically under the plane while falling 3) move ahead of the plane while falling 4) not fall at all

32. Copyright © by Holt, Rinehart and Winston. All rights reserved. Concept Check – Horizontal Projectiles You drop a package from a plane flying at constant speed in a straight line. Without air resistance, the package will: 1) quickly lag behind the plane while falling 2) remain vertically under the plane while falling 3) move ahead of the plane while falling 4) not fall at all same horizontal velocity maintainx-direction Both the plane and the package have the same horizontal velocity at the moment of release. They will maintain this velocity in the x-direction, so they stay aligned. Follow-up: What would happen if air resistance is present?

33. Copyright © by Holt, Rinehart and Winston. All rights reserved. Concept Check – Horizontal Projectiles (2) From the same height (and at the same time), one ball is dropped and another ball is fired horizontally. Which one will hit the ground first? 1) the “dropped” ball 2) the “fired” ball 3) they both hit at the same time 4) it depends on how hard the ball was fired 5) it depends on the initial height

34. Copyright © by Holt, Rinehart and Winston. All rights reserved. Concept Check – Horizontal Projectiles (2) From the same height (and at the same time), one ball is dropped and another ball is fired horizontally. Which one will hit the ground first? 1) the “dropped” ball 2) the “fired” ball 3) they both hit at the same time 4) it depends on how hard the ball was fired 5) it depends on the initial height They both fall from the same height.Therefore, they will hit the ground at the same time. Both of the balls are falling vertically under the influence of gravity. They both fall from the same height. Therefore, they will hit the ground at the same time. The fact that one is moving horizontally is irrelevant – remember that the x and y motions are completely independent !! Follow-up: Is that also true if there is air resistance?

35. Copyright © by Holt, Rinehart and Winston. All rights reserved. Concept Check – Horizontal Projectiles (3) In the previous problem, which ball has the greater velocity at ground level? 1) the “dropped” ball 2) the “fired” ball 3) neither – they both have the same velocity on impact 4) it depends on how hard the ball was thrown

36. Copyright © by Holt, Rinehart and Winston. All rights reserved. Concept Check – Horizontal Projectiles (3) In the previous problem, which ball has the greater velocity at ground level? 1) the “dropped” ball 2) the “fired” ball 3) neither – they both have the same velocity on impact 4) it depends on how hard the ball was thrown same vertical velocity “fired” ball has a larger net velocity Both balls have the same vertical velocity when they hit the ground (since they are both acted on by gravity for the same time). However, the “fired” ball also has a horizontal velocity. When you add the two components vectorially, the “fired” ball has a larger net velocity when it hits the ground. Follow-up: What would you have to do to have them both reach the same final velocity at ground level?

37. Copyright © by Holt, Rinehart and Winston. All rights reserved. Concept Check – Horizontal Projectiles (4) For a cannon on Earth, the cannonball would follow path 2. Instead, if the same cannon were on the Moon, where g = 1.6 m/s 2, which path would the cannonball take in the same situation? 1243

38. Copyright © by Holt, Rinehart and Winston. All rights reserved. Concept Check – Horizontal Projectiles (4) For a cannon on Earth, the cannonball would follow path 2. Instead, if the same cannon were on the Moon, where g = 1.6 m/s 2, which path would the cannonball take in the same situation? 1243 more time farther The ball will spend more time in the air because g Moon < g Earth. With more time, it can travel farther in the horizontal direction. Follow-up: Which path would it take in outer space?

39. Copyright © by Holt, Rinehart and Winston. All rights reserved. Velocity and Acceleration of an Object in Free Fall Velocity Time Acceleration Time Position Time -9.81 m/s 2

40. Copyright © by Holt, Rinehart and Winston. All rights reserved. Projectiles, continued

41. Copyright © by Holt, Rinehart and Winston. All rights reserved. Kinematic Equations for Projectiles Horizontal Component v x = v x,i = constant  x = v x ·  t Vertical Component a y = -g = -9.81 m/s 2 v y,f = v y,i + a y (  t)  y = v y,i (  t) + ½a y (  t) 2 v 2 y,f = v 2 y,i + 2a y (  y)  y = ½(v y,f + v y,f )  t (not included – not necessary)

42. Copyright © by Holt, Rinehart and Winston. All rights reserved. Projectiles Launched Horizontally The initial velocity is the initial horizontal velocity. The initial vertical velocity is 0 ( v y,i = 0 ). Horizontal Component (Constant Velocity) v x = v x,i = constant  x = v x ·  t Vertical Component (Constant Acceleration) a y = -g = -9.81 m/s 2 v y,f = a y (  t)  y = ½a y (  t) 2 v y,f 2 = 2a y (  y) Kinematic Equations for Projectiles

43. Copyright © by Holt, Rinehart and Winston. All rights reserved. Sample Problem

44. Copyright © by Holt, Rinehart and Winston. All rights reserved. Sample Problem

45. Copyright © by Holt, Rinehart and Winston. All rights reserved. Projectiles Launched at an Angle Resolve the initial velocity into x and y components. The initial horizontal velocity is the x- component. The initial vertical velocity is the y- component. Use components in equations Vertical (Constant Acceleration) a y = -g = -9.81 m/s 2 v y,f = v i sin  + a y (  t)  y = (v i sin  t +½a y (  t) 2 v y,f 2 = (v i sin  ) 2 + 2a y (  y) Horizontal (Constant Velocity) v x = v x,i = v i cos  = constant  x = v i cos  ·  t Vertical (Constant Acceleration) a y = -g = -9.81 m/s 2 v y,f = v y,i + a y (  t)  y = v y,i  t +½a y (  t) 2 v y,f 2 = v y,i 2 + 2a y (  y)

46. Copyright © by Holt, Rinehart and Winston. All rights reserved. Upward Moving Projectiles

47. Copyright © by Holt, Rinehart and Winston. All rights reserved. Consider the situation depicted here. A gun is aimed directly at a dangerous criminal hanging from the gutter of a building. The target is well within the gun’s range, but the instant the gun is fired and the bullet moves with a speed v o, the criminal lets go and drops to the ground. What happens? The bullet 1. hits the criminal regardless of the value of v o. 2. hits the criminal only if v o is large enough. 3. misses the criminal. Concept Check – Projectile Motion

48. Copyright © by Holt, Rinehart and Winston. All rights reserved. Consider the situation depicted here. A gun is aimed directly at a dangerous criminal hanging from the gutter of a building. The target is well within the gun’s range, but the instant the gun is fired and the bullet moves with a speed v o, the criminal lets go and drops to the ground. What happens? The bullet 1. hits the criminal regardless of the value of v o. 2. hits the criminal only if v o is large enough. 3. misses the criminal. Concept Check – Projectile Motion Gravity will cause both the bullet and the criminal to accelerate downward at the same rate. Both objects will fall the same distance below their respective zero gravity lines.

49. Copyright © by Holt, Rinehart and Winston. All rights reserved. Concept Check – Projectile Motion (2) A battleship simultaneously fires two shells at enemy ships. If the shells follow the parabolic trajectories shown, which ship gets hit first? 1. A 2. both at the same time 3. B 4. need more information

50. Copyright © by Holt, Rinehart and Winston. All rights reserved. Concept Check – Projectile Motion (2) A battleship simultaneously fires two shells at enemy ships. If the shells follow the parabolic trajectories shown, which ship gets hit first? 1. A 2. both at the same time 3. B 4. need more information The time that a projectile remains in the air depends only on its maximum height. Remember that the vertical component of a projectile’s motion is identical to an object in free fall. The higher it goes, the longer it will remain in the air. Since B does not go as high as A, it will land sooner.

51. Copyright © by Holt, Rinehart and Winston. All rights reserved.

53. Sample Problem

54. Copyright © by Holt, Rinehart and Winston. All rights reserved. Sample Problem

55. Copyright © by Holt, Rinehart and Winston. All rights reserved. Sample Problem

56. Copyright © by Holt, Rinehart and Winston. All rights reserved. Section Review 3-3 1.Parabolic motion is exhibited by projectiles that are under the influence of gravity only. Therefore, examples a,b, and d exhibit parabolic motion.

57. Copyright © by Holt, Rinehart and Winston. All rights reserved. Sample Problem Projectiles Launched At An Angle A zookeeper finds an escaped monkey hanging from a light pole. Aiming her tranquilizer gun at the monkey, she kneels 10.0 m from the light pole,which is 5.00 m high. The tip of her gun is 1.00 m above the ground. At the same moment that the monkey drops a banana, the zookeeper shoots. If the dart travels at 50.0 m/s,will the dart hit the monkey, the banana, or neither one?

58. Copyright © by Holt, Rinehart and Winston. All rights reserved. 1. Select a coordinate system. The positive y-axis points up, and the positive x-axis points along the ground toward the pole. Because the dart leaves the gun at a height of 1.00 m, the vertical distance is 4.00 m. Sample Problem

59. Copyright © by Holt, Rinehart and Winston. All rights reserved. 2. Use the inverse tangent function to find the angle that the initial velocity makes with the x-axis. Sample Problem

60. Copyright © by Holt, Rinehart and Winston. All rights reserved. 3. Choose a kinematic equation to solve for time. Rearrange the equation for motion along the x-axis to isolate the unknown  t, which is the time the dart takes to travel the horizontal distance. Sample Problem

61. Copyright © by Holt, Rinehart and Winston. All rights reserved. 4. Find out how far each object will fall during this time. Use the free-fall kinematic equation in both cases. For the banana, v i = 0. Thus:  y b = ½a y (  t) 2 = ½(–9.81 m/s 2 )(0.215 s) 2 = –0.227 m The dart has an initial vertical component of velocity equal to v i sin , so:  y d = (v i sin  )(  t) + ½a y (  t) 2  y d = (50.0 m/s)(sin  )(0.215 s) +½(–9.81 m/s 2 )(0.215 s) 2  y d = 3.99 m – 0.227 m = 3.76 m Sample Problem

62. Copyright © by Holt, Rinehart and Winston. All rights reserved. 5. Analyze the results. Find the final height of both the banana and the dart. y banana, f = y b,i +  y b = 5.00 m + (–0.227 m) y banana, f = 4.77 m above the ground The dart hits the banana. The slight difference is due to rounding. y dart, f = y d,i +  y d = 1.00 m + 3.76 m y dart, f = 4.76 m above the ground Sample Problem