Physical Quantities are of 2 types…
 A scalar is only a magnitude (length)
(Example: Temperature, time, mass)
 A vector has magnitude and direction
(Example: displacement = 10 m East,
Velocity= 50 mph west)
 A vector will be symbolized by the
“letter” with an arrow over it. The
arrow indicates direction.
A
 Vectors are equal if they have the
same units, magnitude, and
direction.
 A vector can be moved anywhere
parallel to itself.
Adding Vectors (attach)
 To add vectors they must have the same units.
 Tip-to -tail method put them head to tail and connect
them so you end up with a triangle.
 Parallelogram Method- (put them tail to tail) make
vectors parallel and draw a line making 2 triangles
More Properties of Vectors
 Resultant Vector
 The resultant vector is the sum of a given set of
vectors
Subtracting Vectors
 Tip to tail- subtract by
putting vector in the
opposite direction
 If you change the sign of a
vector it is not the same
vector. It is a new vector.
 A – B does not equal
B-A
Components of a Vector
 A component is a part
 It is useful to use
rectangular
components
 These are the
projections of the vector
along the x- and y-axes
Components of a Vector, cont.
 The x-component of a vector is the projection along
the x-axis
Ax  A cos 
 The y-component of a vector is the projection along
the y-axis
 Then,
Ay  A sin 
A  Ax  Ay
Useful Formulas..
A
2
x
2
y
A A
and
 Ay 
  tan  
 Ax 
1
The Pythagorean Theorem can only be used
with right triangles!
When its not 900
R2= A2 + B2 – 2AB(COSӨ)
Example 1
 Find the magnitude of the sum of a 15 km
displacement and a 25 km displacement when the
angle between them is 900 and when the angle
between them is 1350.
Example Problem
(a) Find the horizontal and vertical
components of the 100m
displacement of a superhero who
flies from the top of a tall building
at an angle of 30.00
(b) (b) Suppose instead the superhero
leaps in the other direction along a
displacement vector B to the top
of a flagpole where the
displacement components are
given Bx = -25m and BY=10.0m.
Find the magnitude and direction
of the displacement vector.
Example 2
 A GPS receiver indicates that your home is 15.0 km and
400 north of west, but the only path through the woods
leads directly north. If you follow the path 5.0 km
before it opens into a field, how far, and in what
direction, would you have to walk to reach your
home?
 R= 12.39
 Ө= 158’
Resolving a Vector Into Components
The horizontal, or
x-component, of A is
found by Ax = A cos .
+y
Ay
A

+x
Ax
The vertical, or
y-component, of A is found by Ay = A sin .
By the Pythagorean Theorem, Ax2 + Ay2 = A2.
Every vector can be resolved using these
formulas, such that A is the magnitude of A, and
 is the angle the vector makes with the x-axis.
Each component must have the proper “sign”
according to the quadrant the vector terminates in.
Analytical Method of Vector Addition
1. Find the x- and y-components of each vector.
Ax = A cos  =
Bx = B cos  =
Cx = C cos  =
Ay = A sin  =
By = B sin  =
Cy = C sin  =
Rx =
Ry =
2. Sum the x-components.
This is the x-component of the resultant.
3. Sum the y-components.
This is the y-component of the resultant.
4. Use the Pythagorean Theorem to find the
magnitude of the resultant vector.
Rx2 + Ry2 = R2
 A roller coaster moves 215 ft
horizontally and then rises 130 ft at an
angle of 35.00 above the horizontally.
Next, it travels 125 ft at an angle of
50.00 below the horizontal. Find the
roller coaster’s displacement from its
starting point to the end of this
movement.
You try this one on your own…
Add the following vectors via the component method:
Vector A = 4m South, Vector B= 7.3 m Northwest
 Find R and Ө
 A person walks 25.0° north of east for 2.80
km. How far would the person walk due
north and due east to arrive at the same
location?
 1.183 km North
 2.53 km East
 A person walks East for 255m and then 60° North of
East for 100m. Find the magnitude and direction of the
resultant displacement.
 A vector has an x-componet of -25.0 units and a y-
componet of 45.0 units. Find the magnitude and
direction of the vector.
 A quarter back takes the ball
from the line of scrimmage, runs
backwards for 15.0 yards, then
runs sideways parallel to the line
of scrimmage for 15.0 yards. At
this point, he throws a 60.0 yard
forward pass straight downfield,
perpendicular to the line of
scrimmage. What is the
magnitude of the football’s
resultant displacement?
 A Novice Golfer on the green takes three
strokes to sink the ball. The successive
displacement of the ball are 5.00 m to the
north, 2.00 m 550 north of east, and 1.00 m
at 300 west of south. Starting at the same
initial point, an expert golfer could make
the hole in what singe displacement?
Practice Problems 3A-3C
 Problems 1-4 (pg 89), 1-4 (pg 92),
1-4 pg 94
A projectile is a motion in two dimensions
 Moving in the x and y direction
 A projectile is an object shot through the air.
This occurs in a parabola curve.
projectile- any object that moves through the air
or through space, acted on only by gravity
(and air resistance, if any)
Object
dropped
Object
thrown up
Object thrown at an
angle
-The vertical acceleration of a
projectile is caused by gravity, so
ay = -9.8 m/s2
Parabolic
Trajectory
We assume that
 g remains constant (g= -9.8m/s2)
 a in the x direction is 0 because gravity is not acting on
it.
 Neglect air resistance
 Neglect the effects of the earths rotation
Projectiles launched horizontally
To find how far the ball falls, you
use the formula. Y =viyt + 1/2gt2
1st second- 5m
After 2 seconds- 20m
After 3 seconds- 45m
The curved path of a
projectile produced is a
parabola (caused by both
horizontal motion and vertical
motion. It must accelerate
only in the vertical direction)
A projectile Motion is 2 dimensions
 The projectile will experience two:
 Accelerations (ax= o and aY= -9.8m/s2)
 Velocities
 Displacements
Upwardly Launched Projectiles
When a projectile is launched at an upward
angle, it follows a curved path and finally
hits the ground because of gravity.
The Vertical distance a cannonball falls below “imaginary
path if no gravity” is the same vertical distance it
would fall if it were dropped from rest & had been
falling for the same amount of time.
How to solve these problems
 Draw a free body diagram with a coordinate system.
 Divide the information into x and y components
 Look at your formulas and decided which one(s) to
use.
Formulas for horizontal and vertical motion of a
projectile
(X) Horizontal
xf-xi = vixt + ½ axt 2
vfx = vix + axt
vfx2 - vix2 = 2ax(xf-xi)
(Y) Vertical
yf-yi = viyt + ½ ayt 2
vfy = viy + ayt
vfy2 = viy2 + 2ay (yf-yi)
Objects that have been thrown will have a
horizontal velocity that stays the same (no
horizontal acceleration ax = 0m/s2)
So vfx =vix in the second formula and third
formulas under horizontal motion.
One more formula….
This equation only works when yf and yi are both 0….
t= 2viy
a
If a ball is thrown up in the air from a moving truck, where will it land?
(Ignore air resistance)
In front of the truck, behind the truck, or back in the truck
Where will a package land if it is released
from a plane?
Behind the plane, in front of the plane
below the plane
Look at the example lets
try it together.
 A ball is launched at 4.5m/s at 660 above the
horizontal. What are the maximum height and flight
time of the ball?
 What is the time?
 Maximum height (y)?
 Total displacement (x)?
What is the horizontal distance covered by
an arrow that was shot through the air at a
600 angle with a velocity of 55 m/s?
Given
Solve
v = 55m/s
Vxi = cos 60(55m/s)=27.5 m/s
vix=27.5 m/s
Viy = sin60(55m/s)=47.6m/s
viy=47.6m/s
To find x dist: x = v0x t
ax=0
Need to find time first!
ay=-9.8m/s2
vfy=viy +ayt
t=?
0 = 47.6m/s + -9.8m/s2 t
dx=?
55m/s
dy
600
Vix
dx
dx = Vix t (we need time)
dx = 27.5 m/s(9.7s)
dx = 266.75m
-47.6m/s = -9.8m/st
4.86 s =t
Total time in the air 4.86s x 2 = 9.7s
 Courtney kicks a soccer ball at
rest on level ground giving it an
initial velocity of 7.8 m/s at an
angle of 320
 How long will the ball be in the
air?
 How high will the ball go?
 What will be its range?
A shell is fired from a ship at an angle of 300 above the
horizontal with a speed of 300m/s. If the shell hits the
ocean’s surface in 30.6s, calculate:
(a.)
the distance traveled horizontally
(b.) the maximum height reached by the shell
(c.) the speed of the shell when it reaches its maximum
height
 A Titan Fan stands at the edge of the
bleachers and throws a stone
horizontally over the edge with a
speed of 18.0 m/s. The bleachers are
50.0 m above a flat horizontal
parking lot.
 (a) How long after being released
does the stone strike the parking lot
below the cliff?
 (b) What is the speed in the
horizontal direction when the stone
reaches the parking lot?
 Peter Rowe kicks a soccer ball at rest on level ground
giving it an initial velocity of 13.0m/s at an angle of
400.
 a. How long will the ball be in the air?
 b. How high will the ball go?
 c. What will be its range?
 A brick is thrown upward from the top of a
building at an angle of 250 to the horizontal
and with an initial speed of 15m/s. If the
brick is in flight for 3.0s, how tall is the
building?
 A fireman 50.0m away from a burning
building directs a stream of water from a
ground-level fire hose at an angle of 30.00
above the horizontal. If the speed of the
stream as it leaves the hose is 40.0m/s, at
what height will the stream of water strike
the building?
 (hint) We have to find time first!
 A stone is thrown horizontally at a speed of 5.0m/s
from the top of a cliff that is 78.4m high.
 How long does it take the stone to reach the bottom of
the cliff?
 How far from the base of the cliff does the stone hit the
ground?
 What are the horizontal and vertical components of the
stone’s velocity just before it hits the ground?
 Lucy and her friend are working at an
assembly plant making wooden toy
giraffes. At the end of the line, the
giraffes go horizontally off the edge of
the conveyor belt and fall into a box
below. If the box is 0.6m below the
level of the conveyor belt and 0.4m
away from it, what must be the
horizontal velocity of the giraffes as
they leave the conveyer belt?
 A player kicks a football from ground level with an
initial velocity of 27.0m/s, 30.00 above the horizontal
What is:
 The ball’s hang time?
 The ball’s maximum height?
 The ball’s range?
 The player kicks the ball with the same
speed, but at 600 from the horizontal.
What is the balls hang time, range, and
maximum height?
Practice Problems 3D-3E
 Problems 1-4 (pg 99), 1-4 (pg 101),
1-4 pg 94
Relative Motion
 Frames of Reference
Observers using different frames of reference may
measure different displacements or velocities for an
object in motion
Relative Velocities
the difference between the velocities relative to some
common point
 Relative Motion: Suppose you are on a train platform as the
train rushes through the station without stopping. Someone
on board the train is pitching a ball, throwing it has hard as
they can towards the back of the train. If the train’s speed is
60 mph and the pitcher is capable of throwing at 60 mph,
what is the speed of the ball as you see it from the platform?
Relative Motion
Moving frame of reference
 A boat heading due north
crosses a river with a speed
of 10.0 km/h. The water in
the river has a speed of 5.0
km/h due east.
In general we have vPA
 vPB  vBA
(a) Determine the velocity of the boat.
(b) If the river is 3.0 km wide how long does it take to cross it?
Practice Problems 3F
 Problems 1-4 (pg 105
 Mixed Review 47-62 (pgs111-113