Adding Vectors in Two Dimensions

Transcription

1 Slide 37 / 125 Adding Vectors in Two Dimensions Return to Table of Contents

2 Last year, we learned how to add vectors along a single axis. The example we used was for adding two displacements. Slide 38 / 125 Adding Vectors

3 Last year, we learned how to add vectors along a single axis. The example we used was for adding two displacements. Slide 39 / 125 Adding Vectors 1. Draw the first vector, beginning at the origin, with its tail at the origin.

4 Last year, we learned how to add vectors along a single axis. The example we used was for adding two displacements. 1. Draw the first vector, beginning at the origin, with its tail at the origin. 2. Draw the second vector with its tail at the tip of the first vector. Slide 40 / 125 Adding Vectors

5 Last year, we learned how to add vectors along a single axis. The example we used was for adding two displacements. Slide 41 / 125 Adding Vectors 1. Draw the first vector, beginning at the origin, with its tail at the origin. 2. Draw the second vector with its tail at the tip of the first vector. 3. Draw the Resultant (the answer) from the tail of the first vector to the tip of the last.

6 Slide 42 / 125 Adding Vectors The direction of each vector matters. In this first case, the vector sum of: 5 units to the East plus 2 units to West is 3 units to the East.

7 The direction of each vector matters. For instance, if the second vector had been 2 units to the EAST (not west), we get a different answer. Slide 43 / 125 Adding Vectors In this second case, the vector sum of: 5 units to the East plus 2 units to EAST is 7 units to the East.

8 But how about if the vectors are along different axes. Slide 44 / 125 Adding Vectors in 2-D For instance, let's add vectors of the same magnitude, but along different axes. What is the vector sum of: 5 units East plus 2 units North

9 1. Draw the first vector, beginning at the origin, with its tail at the origin. Slide 45 / 125 Adding Vectors

10 1. Draw the first vector, beginning at the origin, with its tail at the origin. 2. Draw the second vector with its tail at the tip of the first vector. Slide 46 / 125 Adding Vectors

11 1. Draw the first vector, beginning at the origin, with its tail at the origin. Slide 47 / 125 Adding Vectors 2. Draw the second vector with its tail at the tip of the first vector. 3. Draw the Resultant (the answer) from the tail of the first vector to the tip of the last.

12 Slide 48 / 125 Adding Vectors Drawing the Resultant is the same as we did last year. But calculating its magnitude and direction require the use of right triangle mathematics. c a b We know the length of both SIDES of the triangle (a and b), but we need to know the length of the HYPOTENUSE (c).

13 The magnitude of the resultant is equal to the length of the vector. Slide 49 / 125 Magnitude of a Resultant We get the magnitude of the resultant from the Pythagorean Theorem: c 2 = a 2 + b 2 or in this case: R 2 = R 2 = R 2 = 29 R = # (29) = 5.4 units c =? a = 5 units b = 2 units

14 Slide 50 / What is the magnitude of the Resultant of two vectors A and B, if A = 8.0 units north and B = 4.5 units east?

15 Slide 50 (Answer) / What is the magnitude of the Resultant of two vectors A and B, if A = 8.0 units north and B = 4.5 units east? B = 4.5 units Answer A = 8 units [This object is a pull tab]

16 Slide 51 / What is the magnitude of the Resultant of two vectors A and B, if A = 24.0 units east and B = 15.0 units south?

17 Slide 51 (Answer) / What is the magnitude of the Resultant of two vectors A and B, if A = 24.0 units east and B = 15.0 units south?

18 In physics, we say the direction of a vector is equal to the angle # between a chosen axis and the resultant. Slide 52 / 125 Adding Vectors In Kinematics, we will primarily use the x-axis to measure θ. # However, if we were to change the axis we used and apply the proper mathematical techniques, we should get the same result!

19 To find the value of the angle θ, we need to use what we already know: Slide 53 / 125 Adding Vectors the length of the two sides opposite and adjacent to the angle. # (remember SOH CAH TOA)

20 Slide 54 / 125 Adding Vectors tan(θ) = opp / adj tan(θ) = (2 units) / (5 units) tan(θ) = 2/5 tan(θ) = 0.40 # To find the value of θ, we must take the inverse tangent: θ = tan -1 (0.40) = 22 0

21 Slide 55 / 125 Adding Vectors # The Resultant is 5.4 units in the direction of 22 0 North of East magnitude direction

22 Slide 56 / What is the direction of the Resultant of the two vectors A and B if: A = 8.0 units north and B = 4.5 units east if East is 0 o and North is 90 o?

23 Slide 56 (Answer) / What is the direction of the Resultant of the two vectors A and B if: A = 8.0 units north and B = 4.5 units east if East is 0 o and North is 90 o? Answer A = 8 units [This object is a pull tab] B = 4.5 units

24 Slide 57 / What is the direction (from East) of the Resultant of the two vectors A and B if: A = 24.0 units east and B = 15.0 units south?

25 Slide 57 (Answer) / What is the direction (from East) of the Resultant of the two vectors A and B if: A = 24.0 units east and B = 15.0 units south?

26 Slide 58 / Find the magnitude and direction of the resultant of two vectors A and B if: A = 400 units north B = 250 units east Magnitude =?

27 Slide 58 (Answer) / Find the magnitude and direction of the resultant of two vectors A and B if: A = 400 units north B = 250 units east Magnitude =? Answer A = 400 units B = 250 units [This object is a pull tab]

28 Slide 59 / Find the magnitude and direction of the resultant of two vectors A and B if: A = 400 units north B = 250 units east Direction =?

29 Slide 59 (Answer) / Find the magnitude and direction of the resultant of two vectors A and B if: A = 400 units north B = 250 units east B = 250 units Direction =? Answer A = 400 units 32 o East of North [This object is a pull tab]

30 Slide 60 / A student walks a distance of 300 m East, then walks 400 m North. What is the magnitude of the net displacement? A B C D 300 m 400 m 500 m 700 m

31 Slide 60 (Answer) / A student walks a distance of 300 m East, then walks 400 m North. What is the magnitude of the net displacement? A B C D 300 m 400 m 500 m 700 m Answer C [This object is a pull tab]

32 Slide 61 / A student walks a distance of 300 m East, then walks 400 m North. What is the total traveled distance? A B C D 300 m 400 m 500 m 700 m

33 Slide 61 (Answer) / A student walks a distance of 300 m East, then walks 400 m North. What is the total traveled distance? A B C D 300 m 400 m 500 m 700 m Answer D [This object is a pull tab]

34 Slide 62 / Two displacement vectors have magnitudes of 5.0 m and 7.0 m, respectively. When these two vectors are added, the magnitude of the sum: A is 2.0 m. B could be as small as 2.0 m, or as large as 12 m. C is 12 m. D is larger than 12 m.

35 Slide 62 (Answer) / Two displacement vectors have magnitudes of 5.0 m and 7.0 m, respectively. When these two vectors are added, the magnitude of the sum: A is 2.0 m. B could be as small as 2.0 m, or as large as 12 m. C is 12 m. B D is larger than 12 m. Answer [This object is a pull tab]

36 Slide 63 / The resultant of two vectors is the largest when the angle between them is A 0 B 45 C 90 D 180

37 Slide 63 (Answer) / The resultant of two vectors is the largest when the angle between them is A 0 B 45 C 90 D 180 Answer A [This object is a pull tab]

38 Slide 64 / The resultant of two vectors is the smallest when the angle between them is: A 0 B 45 C 90 D 180

39 Slide 64 (Answer) / The resultant of two vectors is the smallest when the angle between them is: A 0 B 45 C 90 D 180 Answer D [This object is a pull tab]

40 Slide 65 / 125 Basic Vector Operations Return to Table of Contents

41 Slide 66 / 125 Adding Vectors Adding Vectors in the opposite order gives the same resultant. V 1 + V 2 = V 2 + V 1 V 1 V 2 V 2 V 1

42 Slide 67 / 125 Adding Vectors Even if the vectors are not at right angles, they can be added graphically by using the "tail to tip" method. The resultant is drawn from the tail of the first vector to the tip of the last vector. V V V 2 + = V 1 V 2 V R V 3

43 Slide 68 / 125 Adding Vectors...and the order does not matter. V 3 V 2 V 3 V 1 V 2 + V 1 + = V R Try it. V 3 V 1 V 2 V R

44 Slide 69 / 125 Subtracting Vectors In order to subtract a vector, we add the negative of that vector. The negative of a vector is defined as that vector in the opposite direction. V 1 - = V 2 V 1 + -V 2 = V R -V 2 V 1

45 Slide 70 / 125 Multiplication of Vectors by Scalars A vector V can be multiplied by a scalar c. The result is a vector cv which has the same direction as V. However, if c is negative, it changes the direction of the vector. V 2V -½V

46 Slide 71 / Which of the following operations will not change a vector? A B C D Translate it parallel to itself Rotate it Multiply it by a constant factor Add a constant vector to it

47 Slide 71 (Answer) / Which of the following operations will not change a vector? A B C D Translate it parallel to itself Rotate it Multiply it by a constant factor Add a constant vector to it Answer A [This object is a pull tab]

48 Slide 72 / 125 Vector Components Return to Table of Contents

49 Slide 73 / 125 Adding Vectors by Components Any vector can be described as the sum as two other vectors called components. These components are chosen perpendicular to each other and can be found using trigonometric functions. y V y V # V x x

50 Slide 74 / 125 Adding Vectors by Components In order to remember the right triangle properties and to better identify the functions, it is often convenient to show these components in different arrangements (notice v y below). y V V y # V x x

51 Slide 75 / If a ball is thrown with a velocity of 25 m/s at an angle of 37 above the horizontal, what is the vertical component of the velocity? A B C D 12 m/s 15 m/s 20 m/s 25 m/s

52 Slide 75 (Answer) / If a ball is thrown with a velocity of 25 m/s at an angle of 37 above the horizontal, what is the vertical component of the velocity? A B C D 12 m/s 15 m/s 20 m/s 25 m/s Answer B [This object is a pull tab]

53 Slide 76 / If a ball is thrown with a velocity of 25 m/s at an angle of 37 above the horizontal, what is the horizontal component of the velocity? A B C D 12 m/s 15 m/s 20 m/s 25 m/s

54 Slide 76 (Answer) / If a ball is thrown with a velocity of 25 m/s at an angle of 37 above the horizontal, what is the horizontal component of the velocity? A B C D 12 m/s 15 m/s 20 m/s 25 m/s Answer C [This object is a pull tab]

55 Slide 77 / If you walk 6.0 km in a straight line in a direction north of east and you end up 2.0 km north and several kilometers east. How many degrees north of east have you walked? A 19 B 45 C 60 D 71

56 Slide 77 (Answer) / If you walk 6.0 km in a straight line in a direction north of east and you end up 2.0 km north and several kilometers east. How many degrees north of east have you walked? A 19 B 45 C 60 D 71 Answer A [This object is a pull tab]

57 Slide 78 / 125 Adding Vectors by Components Using the tip-to-tail method, we can sketch the resultant of any two vectors. y V 1 V 2 But we cannot find the magnitude of 'v', the resultant, since v 1 and v 2 are two-dimensional vectors V x

58 Slide 79 / 125 Adding Vectors by Components We now know how to break v 1 and v 2 into components... y V 2 V 2y V 1 V 2x V 1y V 1x x

59 Slide 80 / 125 Adding Vectors by Components And since the x and y components are one dimensional, they can be added as such. y V 2y V y = v 1y + v 2y V 1y V 1x V 2x V x = v 1x + v 2x x

60 Slide 81 / 125 Adding Vectors by Components 1. Draw a diagram and add the vectors graphically. V 1 V 2

61 y Slide 82 / 125 Adding Vectors by Components 1. Draw a diagram and add the vectors graphically. V 1 V 2 2. Choose x and y axes. x

62 y Slide 83 / 125 Adding Vectors by Components 1. Draw a diagram and add the vectors graphically. V 2y 2. Choose x and y axes. V 2 V 1 V 1y V 2x 3. Resolve each vector into x and y components. V 1x x

63 y Slide 84 / 125 Adding Vectors by Components 1. Draw a diagram and add the vectors graphically. V 2y 2. Choose x and y axes. V 2 V 1 V 1y V 2x 3. Resolve each vector into x and y components. V 1x x 4. Calculate each component. v 1x = v 1 cos(θ 1 ) v 2x = v 2 cos(θ 2 ) v 1y = v 1 sin(θ 1 ) v 2y = v 2 sin(θ 2 )

64 y Slide 85 / 125 Adding Vectors by Components V x 1. Draw a diagram and add the vectors graphically. V 2y 2. Choose x and y axes. V y V 2 V 1 V 1y V 2x 3. Resolve each vector into x and y components. V 1x x 4. Calculate each component. 5. Add the components in each direction.

65 y Slide 86 / 125 Adding Vectors by Components V x 1. Draw a diagram and add the vectors graphically. V V 2y 2. Choose x and y axes. V y V 2 V 1 V 1y V 2x 3. Resolve each vector into x and y components. V 1x x 4. Calculate each component. 5. Add the components in each direction. 6. Find the length and direction of the resultant vector.

66 Slide 87 / 125 Example: Graphically determine the resultant of the following three vector displacements: 1. 24m, 30º north of east 2. 28m, 37º east of north 3. 20m, 50º west of south

67 Slide 88 / 125 Example: Graphically determine the resultant of the following three vector displacements: 1. 24m, 30º north of east 28m 2. 28m, 37º east of north 37 O 24m 3. 20m, 50º west of south 20m 50 O 30 O First, draw the vectors.

68 Slide 89 / 125 Example: Graphically determine the resultant of the following three vector displacements: 1. 24m, 30º north of east 28m 2. 28m, 37º east of north 37 O 24m 3. 20m, 50º west of south 20m 50 O 30 O Next, find the x and y components of each.

69 Slide 90 / 125 Example: Graphically determine the resultant of the following three vector displacements: 1. 24m, 30º north of east 28m 2. 28m, 37º east of north 37 O 24m 3. 20m, 50º west of south 20m 50 O 30 O Next, find the x and y components of each.

70 Slide 91 / 125 Example: Graphically determine the resultant of the following three vector displacements: 1. 24m, 30º north of east 28m 2. 28m, 37º east of north 37 O 24m 3. 20m, 50º west of south 20m 50 O 30 O Next, find the x and y components of each. Note: these are both negative because south and west are negative directions.

71 Slide 92 / 125 Now we can put the components in a chart and solve for the resultant vector. x (m) y (m) d d d To find the magnitude of the resultant, use the Pythagorean theorem. Ʃ To find the direction of the resultant, use inverse tangent.

72 Slide 93 / Graphically determine the magnitude and direction of the resultant of the following three vector displacements: m, 30º north of east m, 37º north of east m, 45 o north of east Magnitude =?

73 Slide 93 (Answer) / 125

74 Slide 94 / Graphically determine the magnitude and direction of the resultant of the following three vector displacements: m, 30º north of east m, 37º north of east m, 45 o north of east Direction =?

75 Slide 94 (Answer) / Graphically determine the magnitude and direction of the resultant of the following three vector displacements: m, 30º north of east m, 37º north of east m, 45 o north of east Direction =? Answer [This object is a pull tab]

76 Slide 95 / Which of the following is an accurate statement? A B C D A vector cannot have a magnitude of zero if one of its components is not zero. The magnitude of a vector can be equal to less than the magnitude of one of its components. If the magnitude of vector A is less than the magnitude of vector B, then the x-component of A must be less than the x-component of B. The magnitude of a vector can be either positive or negative.

77 Slide 95 (Answer) / Which of the following is an accurate statement? A B C D A vector cannot have a magnitude of zero if one of its components is not zero. The magnitude of a vector can be equal to less than the magnitude of one of its components. If the magnitude of vector A is less than the magnitude of vector B, then the x-component [This object is a pull tab] of A must be less than the x-component of B. The magnitude of a vector can be either positive or negative. Answer A

78 Slide 96 / 125 Projectile Motion Return to Table of Contents

79 Slide 97 / 125 Projectile Motion: 1st Type A projectile is an object moving in two dimensions under the influence of Earth's gravity. Its path is a parabola. v y v x v a = g Projectile Motion Vertical fall

80 Slide 98 / 125 Projectile Motion: 1st Type Projectile motion can be understood by analyzing the vertical and horizontal motion separately. The speed in the x-direction is constant. v y v x v a = g Projectile Motion Vertical fall The speed in the y-direction is changing.

81 Slide 99 / 125 A mountain lion leaps horizontally from a 7.5 m high rock with a speed of 4.5 m/s. How far from the base of the rock will he land?

82 Slide 100 / 125 A mountain lion leaps horizontally from a 7.5 m high rock with a speed of 4.5 m/s. How far from the base of the rock will he land? First, determine the time it will take for the lion to reach the ground.

83 Slide 101 / 125 A mountain lion leaps horizontally from a 7.5 m high rock with a speed of 4.5 m/s. How far from the base of the rock will he land? Then, determine how far from the base he will land.

84 Slide 102 / A cannon ball is shot from a cannon at a height of 15 m with a velocity of 20 m/s. How far away will the cannon ball land?

85 Slide 102 (Answer) / A cannon ball is shot from a cannon at a height of 15 m with a velocity of 20 m/s. How far away will the cannon ball land? Answer [This object is a pull tab]

86 Slide 103 / A marble rolls off a table from a height of 0.8 m with a velocity of 3 m/s. Then another marble rolls off the same table with a velocity of 4 m/s. Which values are the same for both marbles? Justify your answer qualitatively, with no equations or calculations. A The final speeds of the marbles. B The time each takes to reach the ground. C The distance from the base of the table where each lan

87 Slide 103 (Answer) / A marble rolls off a table from a height of 0.8 m with a velocity of 3 m/s. Then another marble rolls off the same table with a velocity of 4 m/s. Which values are the same for both marbles? Justify your answer B. qualitatively, Both marbles have with only no a velocity equations or calculations. Answer in the horizontal direction; none in the vertical direction. The time it takes to reach the ground depends A The final speeds of the marbles. only on the initial velocity in the B The time each takes vertical to direction, reach the height ground. of the C The distance from table, the and base the acceleration of the table due to where each lan gravity. All three of these things are the same for both. [This object is a pull tab]

88 Slide 104 / 125 Projectile Motion: 2nd Type If an object is launched at an angle with the horizontal, the analysis is similar except that the initial velocity has a vertical component. v y v v x = v v x v y v v x v y v v x v x v y v

89 Slide 105 / 125 v y v v x = v v x v y v v x v y v v x v x Projectile motion can be described by two kinematics v y v equations: horizontal component: vertical component:

90 Slide 106 / 125 v y v v x = v v x v y v v x v y v v x v x v y v Flying time: Horizontal range: Maximum Height:

91 Slide 107 / Ignoring air resistance, the horizontal component of a projectile's velocity: A B C D is zero. remains constant. continuously increases. continuously decreases.

92 Slide 107 (Answer) / Ignoring air resistance, the horizontal component of a projectile's velocity: A B C D is zero. remains constant. continuously increases. continuously decreases. Answer B [This object is a pull tab]

93 Slide 108 / A ball is thrown with a velocity of 20 m/s at an angle of 60 above the horizontal. What is the horizontal component of its instantaneous velocity at the exact top of its trajectory? A 10 m/s B 17 m/s C 20 m/s D zero

94 Slide 108 (Answer) / A ball is thrown with a velocity of 20 m/s at an angle of 60 above the horizontal. What is the horizontal component of its instantaneous velocity at the exact top of its trajectory? A 10 m/s B 17 m/s C 20 m/s D zero Answer A [This object is a pull tab]

95 Slide 109 / Ignoring air resistance, the magnitude of the horizontal component of a projectile's acceleration: A B C D is zero. remains a non-zero constant. continuously increases. continuously decreases.

96 Slide 109 (Answer) / Ignoring air resistance, the magnitude of the horizontal component of a projectile's acceleration: A B C D is zero. remains a non-zero constant. continuously increases. continuously decreases. Answer A [This object is a pull tab]

97 Slide 110 / At what angle should a water-gun be aimed in order for the water to land with the greatest horizontal range? A 0 B 30 C 45 D 60

98 Slide 110 (Answer) / At what angle should a water-gun be aimed in order for the water to land with the greatest horizontal range? A 0 B 30 C 45 D 60 Answer C [This object is a pull tab]

99 Slide 111 / (Mulitple Answer) An Olympic athlete throws a javelin at six different angles above the horizontal, each with the same speed: 20, 30, 40, 60, 70 and 80. Which two throws cause the javelin to land the same distance away? Be prepared to justify your answer. A 30 and 80 B 20 and 70 C 30 and 70 D 30 and 60

100 Slide 111 (Answer) / (Mulitple Answer) An Olympic athlete throws a javelin at six different angles above the horizontal, each with the same speed: 20, 30, 40, 60, 70 and 80. Which two throws cause the javelin to land the same distance away? Be prepared to justify your answer. A 30 and 80 B 20 and 70 C 30 and 70 D 30 and 60 Answer B and D; since the range equation is v 0 2 sin(2θ)/g, complementary angles will have the same answer [This object is a pull tab]

101 Slide 112 / You are throwing a ball for the second time. If the ball leaves your hand with twice the velocity it had on your first throw, its horizontal range R (compared to your first serve) would be A 1.4 R B R/2 C 2R D 4R

102 Slide 112 (Answer) / You are throwing a ball for the second time. If the ball leaves your hand with twice the velocity it had on your first throw, its horizontal range R (compared to your first serve) would be A 1.4 R B R/2 C 2R D 4R Answer D [This object is a pull tab]

103 Slide 113 / When a football in a field goal attempt reaches its maximum height, how does its speed compare to its initial speed? (Justify your answer.) A B C D It is zero. It is less than its initial speed. It is equal to its initial speed. It is greater than its initial speed.

104 Slide 113 (Answer) / When a football in a field goal attempt reaches its maximum height, how does its speed compare to its initial speed? (Justify your answer.) A B C D It is zero. It is less than its initial speed. It is equal to its initial speed. It is greater than its initial speed. Answer B; the initial velocity is v 0 = (v 0x2 +v 0y2 ) the velocity at the maximum height is just v 0y [This object is a pull tab]

105 Slide 114 / A stone is thrown horizontally from the top of a tower at the same instant a ball is dropped vertically. Which object is traveling faster when it hits the level ground below? (Justify your answer.) A B C D It is impossible to tell from the information given. the stone the ball Neither, since both are traveling at the same speed.

106 Slide 114 (Answer) / A stone is thrown horizontally from the top of a tower at the same instant a ball is dropped vertically. Which object is traveling faster when it hits the level ground below? (Justify your answer.) A B C D It is impossible to tell from the information given. the stone the ball Neither, since both are B; both traveling the stone at the and same the speed. Answer ball have will have the same vertical component of the velocity when they hit the ground but the stone will also have a horizontal component [This object is a pull tab]

107 Slide 115 / A plane flying horizontally at a speed of 50.0 m/s and at an elevation of 160 m drops a package. Two seconds later it drops a second package. How far apart will the two packages land on the ground? A B C D 100 m 170 m 180 m 210 m

108 Slide 115 (Answer) / A plane flying horizontally at a speed of 50.0 m/s and at an elevation of 160 m drops a package. Two seconds later it drops a second package. How far apart will the two packages land on the ground? A B C D 100 m 170 m 180 m 210 m Answer A [This object is a pull tab]

109 Slide 116 / An arrow is fired at an angle of θ above the horizontal with a speed of v. Explain how you can calculate the horizontal range and maximum height. (List the steps you would take and the equations you would use so that another student could solve this problem but do not solve the problem.)

110 Slide 116 (Answer) / An arrow is fired at an angle of θ above the horizontal with a speed of v. Explain how you can calculate the horizontal range and maximum height. (List the steps you would take and the Find the vertical and horizontal equations you would use so that another student components of the initial velocity. could solve this problem but do not solve the problem.) With the vertical component, use v=v 0 +at Answer to find the time in the air. With the vertical component and 1/2 the time in the air, use y=y 0 +v oy t+1/2at 2 to find the maximum height. With the horizontal component and the total time in the air, use x=v ox t to find the range. [This object is a pull tab]

111 Slide 117 / A projectile is fired with an initial speed of 30 m/s at an angle of 30 above the horizontal. Determine the total time in the air.

112 Slide 117 (Answer) / A projectile is fired with an initial speed of 30 m/s at an angle of 30 above the horizontal. Determine the total time in the air. Answer [This object is a pull tab]

113 Slide 118 / A projectile is fired with an initial speed of 30 m/s at an angle of 30 above the horizontal. Determine the maximum height reached by the projectile.

114 Slide 118 (Answer) / A projectile is fired with an initial speed of 30 m/s at an angle of 30 above the horizontal. Determine the maximum height reached by the projectile. Answer [This object is a pull tab]

115 Slide 119 / A projectile is fired with an initial speed of 30 m/s at an angle of 30 above the horizontal. Determine the maximum horizontal distance covered by the projectile.

116 Slide 119 (Answer) / A projectile is fired with an initial speed of 30 m/s at an angle of 30 above the horizontal. Determine the maximum horizontal distance covered by the projectile. Answer [This object is a pull tab]

117 Slide 120 / A projectile is fired with an initial speed of 30 m/s at an angle of 30 above the horizontal. Determine the velocity of the projectile 2s after firing.

118 Slide 120 (Answer) / A projectile is fired with an initial speed of 30 m/s at an angle of 30 above the horizontal. Determine the velocity of the projectile 2s after v x firing. v v y Answer [This object is a pull tab]

119 Slide 121 / 125 If an object is launched at an angle with the horizontal and from an initial height, the analysis is similar except when finding the total time in the air. You will need to use the quadratic formula. Projectile Motion: 3rd Type v y v v y v x v v x v x = v v y v x v v y v x v v x v y v

120 Slide 122 / A projectile is fired from the edge of a cliff 200 m high with an initial speed of 30 m/s at an angle of 45 above the horizontal. Determine the total time in the air.

121 Slide 122 (Answer) / A projectile is fired from the edge of a cliff 200 m high with an initial speed of 30 m/s at an angle of 45 above the horizontal. Determine the total time in the air. Answer [This object is a pull tab]

122 Slide 123 / A projectile is fired from the edge of a cliff 200 m high with an initial speed of 30 m/s at an angle of 45 above the horizontal. Determine the maximum horizontal range.

123 Slide 123 (Answer) / A projectile is fired from the edge of a cliff 200 m high with an initial speed of 30 m/s at an angle of 45 above the horizontal. Determine the maximum horizontal range. Answer [This object is a pull tab]

124 Slide 124 / A projectile is fired from the edge of a cliff 200 m high with an initial speed of 30 m/s at an angle of 45 above the horizontal. Determine the magnitude of the velocity just before impact.

125 Slide 124 (Answer) / A projectile is fired from the edge of a cliff 200 m high with an initial speed of 30 m/s at an angle of 45 above the horizontal. Determine the magnitude of the velocity just before impact. Answer [This object is a pull tab]

126 Slide 125 / A projectile is fired from the edge of a cliff 200 m high with an initial speed of 30 m/s at an angle of 45 above the horizontal. Determine the angle the velocity makes with the horizontal just before impact.

127 Slide 125 (Answer) / A projectile is fired from the edge of a cliff 200 m high with an initial speed of 30 m/s at an angle of 45 above the horizontal. Determine the angle the velocity makes with the horizontal just before impact. Answer [This object is a pull tab]