1. Two Dimensional Motion and Vectors
Vector Review
Vector Analysis (Analytical)
Vector Analysis (Graphical)

2. 5-1 Vectors and Scalars A vector has magnitude as well as direction.
Some vector quantities: displacement, velocity, force, momentum
A scalar has only a magnitude.
Some scalar quantities: mass, time, temperature

3. Vectors can be drawn as arrows that are proportional to the magnitude
Vectors can be drawn as arrows that are proportional to the magnitude. This is the graphical representation of a vector.

4. Vectors can be added 2 ways.
1. Graphically method (aka the “head-to-tail” method)
2. component method
The sum of adding 2/more vectors is called the resolution vector.

5. It does not matter what order you add or subtract vectors, the resultant vector will be the same
A = 10 km and B = 20 km Both are acting to the right
Vectors are drawn starting at a point (tail) and ending at an arrow pointing in the direction the vector is acting (head)
Resultant Vector
A = 10 km and B = 20 km A is acting to the right, B left
Subtracting two vectors is the same as adding the opposite vector from above

6. Physically Adding Vectors
In the below left diagram, we see 3 vectors with their associated magnitudes and angles. In order to add these, we always must connect vectors 'head to tail' and the resultant vector (which represents the vector sum) is drawn from the tail of the first vector to the head of the last vector.
In this example, using a ruler and protractor, we are able to get a resultant vector of about magnitude 11 and an angle of 102°.

7. Finding the Magnitude of the Resultant of Perpendicular Vectorsy q x
Use a ruler and the scale you used to produce the other vectors to find the magnitude of the Resultant
Graphically – You measure the angle with a protractor.

8. Vector Operations
Choosing a way to describe direction – Pick an appropriate coordinate system x
One coordinate system is the Cartesian y -y -x
x,y
Angle from x axis

9. Direction can be described by Compass pointsNW NE SW SE
Direction can be described by Compass points
The angle from the x axis (or y axis) is 45O
The direction could also be described as 45O north of east

10. Finding the Magnitude of Perpendicular Vectors
Using graphical methods of finding the magnitude is only an approximation
Using mathematical relationships is a more accurate way to find the magnitude of perpendicular vectors

11. Finding the magnitude of displacement using the Pythagorean Theorem

12. Using Trig Identities – Finding Direction

14. Great resource 

15. 5-6 Adding Vectors by Components
If the components are perpendicular, they can be found using trigonometric functions.

16. Trig functions… Cos θ adjacent side Ax hypotenuse A
Sin θ opposite side Ay hypotenuse A A Ay Ax

17. Find Vr if its components are 15 N and 6E.
15 m/s
6 m/s
Find Vr if its components are 15 N and 6E.
Vr will be hypotenuse, so magnitude will be found Vr2 = = m/s
What will be the resulting angle?

18. tan Θ = opp/adj
Θ = opp/adj
tan
Θ = arctan 15/6 = 68°
Vr = 68°

19. If 1 cm is equal to 20km/hr and
Vector addition
If 1 cm is equal to 20km/hr and
you draw the vectors, the north side is 5 cm and the east is 1 cm the resultant is cm.
100km/hr north
20 km/hr east

20. Resolving Vectors Into Components
Vectors can be described by a set of INDEPENDENT perpendicular components with their proper units
The “x” component lies parallel to the x axis
The “y” component lies parallel to the y axis
The components are call “projections” of the vector on the x and y axis.
The x component has a magnitude of (x,0) and the y component has a magnitude of (0,y)

21. y
(+,+) I
(-,+) II x
(+,-) IV
(-,-)
III

23. Resolving Vector A into x and y component directions
Ay = sinq A q
(x,0) x
Vector A is the hypotenuse of the right triangle of its x and y components
Ax = cosq A

24. Adding Non-Perpendicular Vectors Algebraically
Review – Vectors that are perpendicular (make a right angle with each other) can be added using the Pythagorean Theorem. This gives the magnitude of the resultant vector c.

25. Using Trig to find components next easiest application.
What are the x and y components of a resultant vector 17.5 m/s at 40° using sin and cos we can find the components
sin 40 = Vy cos 40 = Vx
(0.6428)(17.5) = N = y
(0.7660)(17.5) = E = x

26. Find the resultant of a 120N force acting at a 30° east of north and a
90N force acting at 50° east of south.
BOTH vectors must be resolved into H & V

27. Top triangle:
Sin 30 = x1 / 120
Cos 30 = y1 / 120
.5 = x1 / 120
.866 = y1 / 120
x1 = 60N East
y1 = 104N North
Bottom triangle:
Sin 50 = x2 / 90
Cos 50 = y2 / 90
.766 = x2 / 90
.643 = y2 / 90
x2 = 69N East
y2 = 58N South
R2 = =
R2 = 18757
R = √18757 = 137
Tan θ =
opp adj =
= .357
θ = 20
R = 1.4 × 102 N 20° north of east (2 s.f.)

30. Use the method of rectangular resolution to find the resultant of the
following set of forces: 1) 200 N to the right;
2) above the horizontal to the right.
3) above the horizontal to the left;
4) 200 N vertically downward. Suggestion:
vectors
x-components
y-components F1 F2 F3 F4
Fnet

31. The net force can be stated as: 279. 3 N to the right (x) & 130
The net force can be stated as: N to the right (x) & N upward (y). We can also determine the magnitude and direction.
| Fnet | = = = N.
The direction can be determined by redrawing the net force with it two components. From the figure we then have:
tan  = (130.5(/(279.3) = .467   = 250.
In words: The net force = above the horizontal to the right.

33. 100 km 150 km 200 km 300 km none of the above
4. You fly east in an airplane for 100 km. You then turn left 60 degrees and travel 200 km. How far east of the starting point are you? (approximately)
100 km
150 km
200 km
300 km
none of the above
Adapted from Beale/Franklin
C   200 km.
x=100 km + (200 km)cos(60°)=100 km + 100km = 200 km

34. x=100 km + (200 km)cos(60°)=100 km + 100km = 200 km
C   200 km.
x=100 km + (200 km)cos(60°)=100 km + 100km = 200 km

38. And here is how to add two vectors after breaking them into x and y parts:
The vector (8,13) and the vector (26,7) add up to the vector (34,20)
Example: add the vectors a = (8,13) and b = (26,7)
c = a + b
c = (8,13) + (26,7) = (8+26,13+7) = (34,20)

39. Let us add the two vectors head to tail:
An Example
Sam and Alex are pulling a box.
Sam pulls with 200 Newtons of force at 60°
Alex pulls with 120 Newtons of force at 45° as shown
What is the combined force, and its direction?

40. Let us add the two vectors head to tail:
Now, convert from polar to Cartesian (to 2 decimals):
Sam's Vector:
x = r × cos( θ ) = 200 × cos(60°) = 200 × 0.5 = 100
y = r × sin( θ ) = 200 × sin(60°) = 200 × =
Alex's Vector:
x = r × cos( θ ) = 120 × cos(-45°) = 120 × = 84.85
y = r × sin( θ ) = 120 × sin(-45°) = 120 × =

41. Now we have:
Now it is easy to add them:
(100, ) + (84.85, ) = (184.85, 88.36)

42. We can convert that to polar for a final answer:
r = √ ( x2 + y2 ) = √ ( ) =
θ = tan-1 ( y / x ) = tan-1 ( / ) = 25.5°
                                          
And we have this (rounded) result:
And it looks like this for Sam and Alex:

43. Non-Perpendicular Vectors can be added by finding the x and y components of the vectors and adding or subtracting the one dimensional components Organizing these problems:
Determine an appropriate coordinate system for the vectors
Label the vectors
Resolve all vectors into their x and y components
List the x and y components of the vectors
Add all of the x components of the vectors and all of the y components of the vectors
The above yields the x and y components of the resultant vector

44. The y component of vector A – the y component of vector B = the y component of the resultant vector
The x component of vector A + the x component of vector B = the x component of the resultant vector

45. Adding Two Dimensional Vectorsy
Remember, it does not matter what order you add the vectors, the resultant will be the same!
Resultant D C B
The direction of the Resultant is from the x axis q A X
Draw the resultant vector from the tail of the first vector in the direction of the head of the last vector. The magnitude will be the length of the Resultant Vector.

46. A plane flies 275 mi/hr at 215. 5° while a wind blows 95. 5 mi/hr west
A plane flies 275 mi/hr at 215.5° while a wind blows 95.5 mi/hr west. Find the resultant velocity of the plane after 1 hour.
Vpx = (cos 215.5)(275mi/hr) = mi/hr
-224 mi/hr x or 224 mi/hr W
Vpy = (sin 215.5)(275mi/hr) = mi/hr
-160 mi/hr y or 160 mi/hr S
Vw = 95.5 W
Vx= 224 W W = W; Vy= 160 S
Vr = 357 mi/hr at 207°

47. A laser beam is aimed 15.95° above the horizontal at a mirror 11,648 m away. It glances off the mirror and continues for an additional 8570. m at 11.44° above the horizon until it hits its target. What is the resultant displacement of the beam to the target?

48. A laser beam is aimed 15.95° above the horizontal at a mirror 11,648 m away. It glances off the mirror and continues for an additional 8570. m at 11.44° above the horizon until it hits its target. What is the resultant displacement of the beam to the target?
Vectors that are not at nice angles need to be dealt with. Break them up into their components.

49. If the two vectors are NOT perpendicular to each other.
LAW OF SINES AND
LAW OF COSINES
Looking for side c use the law of cosines:
Looking for <A use the law of sines.(<A is the smallest of the angles) c a q A b

50. Notice: the vectors are tail to tip.
A man walks 10 m east then turns and walks N of E. Find the resultant displacement.
Notice: the vectors are tail to tip. d
15 m
350
10 m

51. We need to find this angle.
A man walks 10 m east then turns and walks N of E. Find the resultant displacement.
= 1450
1450
We need to find this angle.
Now for the angle. The angle is the one from the original vector to the resultant vector. Always measured from the starting vector. (motion problems) d
15 m
350 A
10 m
We know that <A is acute so the law of sines is okay.
Displacement: N of E

52. Anything that is thrown or shot through the air.
What is a Projectile?
Anything that is thrown or shot through the air.
Projectiles have velocities in two directions.
Horizontal Motion: Motion parallel to the Earth’s surface.
Vertical Motion: The force of gravity pulling down on the object.

53. Horizontal Launched Projectiles
Horizontally launched projectiles have an initial Vy of zero
The objects velocity Vx is constant (gravity only affects objects that are moving up or down)
Projectile motion is “free fall” motion with an initial Vx

54. Projectile Motion
Objects that are thrown or launched into the air are accelerated by gravity and are projectiles
In our problems, projectiles will ONLY be affected by gravity. The initial force to launch, air resistance while in flight, and the force used to stop the projectile are not considered part of the motion of the projectile
The path of the projectile is parabolic and symmetrical
Analyzing Projectile Motion is an application of vector components

55. When objects fall, the Horizontal Velocity does not affect the Vertical Component because they are Perpendicular.
If the purple ball is dropped at the same time the yellow ball falls off of the table ….. Vx Vx
…..the objects fall at the same rate even though the purple ball has a Horizontal Velocity Component Vy Vy
They both will hit the floor at the same time.

56. WHAT VARIABLE IS THE SAME IN HORIZONTAL AND VERTICAL VELOCITY OF A PROJECTILE?

57. Projectiles…..

58. Projectile Motion
It can be understood by analyzing the horizontal and vertical motions separately.

59. Projectile Motion
The speed in the x-direction is constant; in the y-direction the object moves with constant acceleration g.
This photograph shows two balls that start to fall at the same time. The one on the right has an initial speed in the x-direction. It can be seen that vertical positions of the two balls are identical at identical times, while the horizontal position of the yellow ball increases linearly.

60. Vx Vy Vy VT
The total velocity is the vector sum of the perpendicular velocities Vx and Vy
VT =√ (Vx2 + Vy2)
The total velocity of this ball is just Vy at any time

61. Any of these formulas can be used to solve the projectile motion problems
VF 2 = Vo a(Δd)
Δd = Vo Δt ½ a (t2)
Vf = Vo + aΔt

63. The common variable of the Horizontal and Vertical motion is Time
The common variable of the Horizontal and Vertical motion is Time. (remember the pings of the 2 marbles)
The object will travel horizontally and vertically for the same time interval
If you are given the time interval (or can calculate the time) in one direction, you can use that time to find unknowns in BOTH directions
tv = th
tv = th
tv = th
tv = th
tv = th
tv = th
tv = th

64. Projectiles Launched at an Angle
The initial vertical velocity, Vy of the projectile is not zero
The velocity in the x direction, Vx is constant for the entire flight
The total velocity, VT of the projectile is the vector sum of the perpendicular Vx and Vy velocities (at any point in the projectile’s path)

66. Analyzing Projectile Motion With Velocity Vectors
After the cannon ball is launched, the ball’s Horizontal Velocity Component remains constant. No force is acting in the horizontal direction. Notice the path of the ball is SYMETRICAL with no air resistance. This is important to remember when solving problems. Vy Vx
However, gravity decreases the Vertical Component as it rises until it is 0, and it increases in the downward direction as it falls.

68. Practice Problem 1
In her Physics lab, Melanie rolls a 10g marble down a ramp and off the table with a horizontal velocity of 1.2 m/s. The marble falls in a cup placed .51 m from the table’s edge. How high is the table?

69. And the answer is
First find the time in the air (same time for vertical and horizontal)
t = dx / vx
.51m /1.2m/s = .43s
Since things fall at the same rate…..you can use the same time to find the vertical distance it traveled.
Δdy = vyΔt ½ gt2
Δdy = ½ (-9.8 m/s2) (.43s)2
= -.9 m

70. First find horizontal and vertical displacements:
Example 2: A baseball is hit with a horizontal speed of 25 m/s. What is its position and velocity after 2 s?
25 m/s x y
-19.6 m
+50 m
First find horizontal and vertical displacements:
x = 50.0 m
y = m

71. Example 1 Cont.): What are the velocity components after 2 s?
25 m/s vx vy
v0x = 25 m/s v0y = 0
Find horizontal and vertical velocity after 2 s:
vx = 25.0 m/s
vy = m/s

72. Consider Projectile at an Angle:
A red ball is projected at an angle q. At the same time, a yellow ball is thrown vertically upward and a green ball rolls horizontally (no friction). q
vx = vox = constant
voy
vox vo
Note vertical and horizontal motions of balls

73. Displacement Calculations For General Projection:
The components of displacement at time t are:
For projectiles:
Thus, the displacement components x and y for projectiles are:

74. Velocity Calculations For General Projection:
The components of velocity at time t are:
For projectiles:
Thus, the velocity components vx and vy for projectiles are:

75. Problem-Solving Strategy:
Resolve initial velocity vo into components: vo
vox
voy q
2. Find components of final position and velocity:
Displacement:
Velocity:

76. Example 2: A ball has an initial velocity of 160 ft/s at an angle of 30o with horizontal. Find its X position and velocity after 2 s and after 4 s.
voy
160 ft/s
30o
vox
Since vx is constant, the horizontal displacements after 2 and 4 seconds are:
x = 277 ft
x = 554 ft

77. Example 2: (Continued)
2 s
4 s
voy
160 ft/s
vox
30o
277 ft
554 ft
Note: We know ONLY the horizontal location after 2 and 4 s. We don’t know whether it is on its way up or on its way down.
x2 = 277 ft
x4 = 554 ft

78. Example 2 (Cont.): Next we find the vertical components of position after 2 s and after 4 s.
voy= 80 ft/s
160 ft/s q
0 s
3 s
2 s
1 s
4 s
g = -32 ft/s2 y2 y4
The vertical displacement as function of time:
Observe consistent units.

79. (Cont.) Signs of y will indicate location of displacement (above + or below – origin).
voy= 80 ft/s
160 ft/s q
0 s
3 s
2 s
1 s
4 s
g = -32 ft/s2 y2 y4
96 ft
16 ft
Vertical position:
Each above origin (+)

80. Vertical velocity is same as if vertically projected:
(Cont.): Next we find horizontal and vertical components of velocity after 2 and 4 s.
voy
160 ft/s
vox
30o
Since vx is constant, vx = 139 ft/s at all times.
Vertical velocity is same as if vertically projected:
At any time t:

81. Example 2: (Continued) g = -32 ft/s2 160 ft/s q vy= 80.0 ft/s v2 v4
At any time t:
v2y = 16.0 ft/s
v4y = ft/s

82. Example 2: (Continued) g = -32 ft/s2 160 ft/s q Moving Up +16 ft/s
vy= 80.0 ft/s
160 ft/s q
0 s
3 s
2 s
1 s
4 s
g = -32 ft/s2 v2 v4
Moving Up +16 ft/s
Moving down -48 ft/s
The signs of vy indicate whether motion is up (+) or down (-) at any time t.
At 2 s: v2x = 139 ft/s; v2y = ft/s
At 4 s: v4x = 139 ft/s; v4y = ft/s

83. (Cont.): The displacement R2,q is found from the x2 and y2 component displacements.
y2 = 96 ft q
x2= 277 ft
0 s
2 s
4 s
R2 = 293 ft
q2 = 19.10

84. (Cont.): Similarly, displacement R4,q is found from the x4 and y4 component displacements.
y4 = 64 ft
x4= 554 ft R4
t = 4 s
R4 = 558 ft
q4 = 6.590

85. (Cont.): Now we find the velocity after 2 s from the components vx and vy.
voy= 80.0 ft/s
160 ft/s q
0 s
2 s
g = -32 ft/s2 v2
Moving Up +16 ft/s
v2x = 139 ft/s
v2y = ft/s
v2 = 140 ft/s
q2 = 6.560

86. (Cont.) Next, we find the velocity after 4 s from the components v4x and v4y.
voy= 80.0 ft/s
160 ft/s q
0 s
4 s
g = -32 ft/s2 v4
v4x = 139 ft/s
v4y = ft/s
v4 = 146 ft/s
q2 =

87. An arrow is shot at a 30. 0° angle above horizontal ground
An arrow is shot at a 30.0° angle above horizontal ground.  It has an initial speed of 49 m/s.  (a) How high will the arrow go?  (b) What horizontal distance will it travel?
a. 31 m
b. 210 m

88. 2 D Projectile Motion
The kinematics equations mentioned in the previous
slide are as follows:
1. vix = vi (cos θ) = constant (x direction, a = 0)
2. dx = vi (cos θ) t (x direction, a = 0)
3. vyf = vi (sin θ) + a t (y direction, a = g)
4. vyf 2 = vi 2 (sin θ) a dy (y direction, a = g)
5. dy = vi (sin θ) t + ½ a t (y direction, a = g)

89. 2 D Projectile Motion Example One (projectile launched at an angle):
A projectile is launched at 50 m/s at an angle of 45o
from the horizontal. It lands at a place level with the
place that it was launched from. Find:
a. time of flight of the projectile (t total = ?)
b. maximum height of the projectile (dy = ?)
c. range of the projectile (dx = ?)
vi = 50 m/s, θ = 45o
given: a = – 9.8 m/s2
vi = 50 m/s
viy = vi (sin θ) θ
θ = 45o
vix = vi (cos θ)

90. 2 D Projectile Motion a. time of flight (t)
dy = vi (sin θ) t + ½ a t 2 (know that dy = 0)
– vi (sin θ) t = ½ a t (dy = displacement)
– vi (sin θ) = ½ a t
b. maximum height (dy)
dy = vi (sin θ) t + ½ a t 2 (use t / 2)
= (50 m/s)(.7071)(3.6 s) + (½)(– 9.8 m/s 2)(3.6 s) 2
= 63.7 m = dy
could have used vyf 2 = vyi 2 (sin θ) a dy
– 2(vi sin θ) a
– 2(50 m/s)(.7071)
– 9.8 m/s2
t = =
= 7.2 s = t

91. 2 D Projectile Motion c. Range (dx) = ?
dx = vx t (substitute: vx = vi cos θ)
dx = (vi cos θ) t
dx = (50 m/s)(.7071)(7.2 s)
dx = m

92. 2 D Projectile Motion Example Two (projectile launched at an angle):
A cat jumps off of a table that is 1 m high. The initial
velocity of the cat is 3 m/s at an angle of 37o above the
horizontal. How far out from the edge of the table does
the cat strike the floor?
given: dy = – 1 m, θ = 37o, vi = 3 m/s find: dx = ?
vi = 3m/s θ dy dx

93. 2 D Projectile Motion
1. First, find the horizontal and vertical components of
the initial velocity.
viy = vi (sin 37o) = 3 m/s (sin 37o) = 1.8 m/s = viy
vix = vi (cos 37o) = 3 m/s (cos 37o) = 2.4 m/s = vix
2. Use this equation to solve for t because t is the only
unknown.
dy = vi (sin θ) t + ½ a t 2
Since dy is not = 0, the only way to solve for t is to use
the quadratic equation.
a. ½ a t 2 + viy t + – dy = 0

94. 2 D Projectile Motion a. ½ a t 2 + viy t + – dy = 0 b.
d. t = 0.67 seconds
(½ (– 9.8 m/s2)) t 2 + (1.8 m/s) t + (+ 1m) = 0
(a) (b) (c)
– b +/– √b 2 – 4 a c
2 a
(quadratic equation)
– 1.8 m/s +/– √(1.8 m/s) 2 – {4(– 4.9 m/s 2)(+1 m)}
2(– 4.9 m/s 2)

95. 2 D Projectile Motion
Then plug t into this equation to finally find dx.
dx = vi (cos θ) t
dx = (2.4 m/s)(0.67 s)
dx = 1.6 m

96. A daredevil tries to jump a canyon of width 10 m
A daredevil tries to jump a canyon of width 10 m. To do so, he drives his motorcycle up an incline sloped at an angle of 15 degrees. What minimum speed is necessary to clear the canyon?

97. vx = Dx / Dt = V cos15o
vy = V sin15o
V = 10m / (.73s) * cos15o) V = m/s, the minimum speed needed for success.

98. Now use the y-component information to find the height:
A brick is thrown upward from the top of a building at an angle of 25 degrees above the horizontal and with an initial speed of 15 m/s. If the brick is in the air for 3 seconds, how high is the building?
Find the initial vertical velocity:
sin 25o = Vy/15m/s Vy = 6.3 m/s
Now use the y-component information to find the height:
Dy = (.5)(g)(t2) + (Vy (t) Dy = (.5)(-9.8m/s2)(3s)2+(6.3m/s)(3s) Dy = -45m +18.9m Dy = -25.2m

99. A brick is thrown upward at an angle of 25 degrees above the horizontal and with an initial speed of 15 m/s. How FAR in the horizontal direction did the brick land from the where it was released?
Cos 25° (15) = Vx
13.6 = Vx
Sin 25° (15) = Vy
Vy = 6.3
Vf = V0 + g t
0= (-9.8) t
T = .65 sec for the time UP so the total time in the air is 1.3 seconds. 
vX = dx/t
13.6 = dx/1.3
dx = m

100. An arrow is shot at a 30. 0° angle above horizontal ground
An arrow is shot at a 30.0° angle above horizontal ground.  It has an initial speed of 49 m/s.  (a) How high will the arrow go?  (b) What horizontal distance will it travel?
Find Vx and Vy
Cos 30 (49)= 42.4= Vx
Sin 30 (49) 24.5= Vy
VF 2 = Vo a(Δd)
0 = (-9.8) d
Dy=

101. Now part B knowing that Vx= 42.4 and we are looking for dx.
Vf = V0 + g t
0= (-9.8) t
T = 2.5 sec for the time UP so the total time in the air is 5 seconds. 
Now solve for dx
Vx = dx/ t
42.4 = dx/5
Dx= 212 m

102. Billy jumps off a high diving platform with a horizontal velocity of 2
Billy jumps off a high diving platform with a horizontal velocity of 2.06 m/s and lands in the water 1.3 seconds later. How high is the platform?
8.21 m
A model rocket flies horizontally off the edge of a cliff at a velocity of 50.0 m/s. If the canyon below is m deep, how far from the edge of the cliff does the model rocket land?
225 m
A stone is thrown horizontally from the top of a m cliff. The stone lands at a distance of m from the edge of the cliff. What is the time needed to hit the ground and the initial horizontal velocity of the stone?
2.9 sec, 8.75 m/s
A boy kicks a football with an initial horizontal velocity of 14 m/s and 7 m/s initial vertical velocity. What is the highest elevation reached by the football in its trajectory?
2.5 m

103. A ball rolls over the edge of a table with a horizontal velocity in m/s. The height of the table is 1.6 m and the horizontal range of the ball from the base of the table is 20 m. What is the initial speed of the ball?
35 m/s