Module 1
1
Scalar and Vector Quantities
q Scalar quantity – quantity that can only be described by a
magnitude (value + unit)
q Vector quantity – quantity that is both described by
magnitude (value + unit) and direction
q
Scalars:
q
q
q
q
q
q
Distance
Speed (magnitude of
velocity)
Temperature
Energy
Mass
Time
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q
Vectors
q
q
q
q
q
q
Displacement
Velocity (magnitude and
direction!)
Acceleration
Force
Weight
Momentum
2
Vector Notation
q
To describe vectors we will use:
n
n
n
n
An arrow above the vector: 𝑨
To describe the magnitude of a
vector we will use absolute
value sign: 𝑨 or just A,
Magnitude is always positive,
the magnitude of a vector is
equal to the length of a vector.
Arrows point the direction
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3
Properties of Vectors
!
A
Consider vectors 𝑨 and 𝑩
1. 𝑨 and 𝑩 are parallel if they don’t intersect and
have the same direction regardless of their
magnitude
2. 𝑨 and 𝑩 are antiparallel if they don’t intersect
but they are in opposite direction regardless of
their magnitude
!
B
!
A
!
B
!
A
3. 𝑨 and 𝑩 are equal (𝑨 = 𝑩) if they are parallel
(same direction) and have the same magnitude
𝑨 = 𝑩
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!
B
4
Properties of Vectors
!
A
4. 𝑨 is a negative vector of 𝑩 (𝑨 = −𝑩) if they
are antiparallel (opposite direction) but have !
the same magnitude 𝑨 = 𝑩
B
5. 𝑐𝑨 is a multiple of 𝑨 by some factor of scalar
c.
!
cA
!
A
!
cA
§ If c is positive, 𝑨 and 𝑐𝑨 are in the same
direction
§ If c is negative, 𝑨 and 𝑐𝑨 are in the
opposite direction
§ If |c|<1, 𝑐𝑨 < 𝑨
!
A
!
cA
!
cA
§ If |c|>1, 𝑐𝑨 > 𝑨
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5
Vector Addition
ü In adding vectors, let 𝑹 be the result
vector or the vector sum of 𝑨 and 𝑩.
That is, 𝑹 = 𝑨 + 𝑩.
(𝑹 is the resultant vector)
𝑩
𝑨
𝑹
ü Note: Adding vectors is not the same or as easy as adding scalars.
Properties of Vector Addition
1) 𝐴⃗ + 𝐵 = 𝐵 + 𝐴⃗
(Commutative)
2) 𝐴⃗ + 𝐵 + 𝐶⃗ = 𝐴⃗ + 𝐵 + 𝐶⃗
(Associative)
3) 𝐴⃗ − 𝐵 = 𝐴⃗ + −𝐵
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Addition/Subtraction of
Vectors
A. Graphical Methods
1. Tail-to-tip or Polygon Method
2. Parallelogram Method
B. Component Method
𝑩
𝑨
𝑹
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Tail-to-tip/Polygon Method
(Triangle Method)
!
A
+
!
B
=
! !
A+ B
Solution:
1. Draw the first vector 𝑨 (with the
appropriate length and direction).
2. Draw the next vector 𝑩 (with the
appropriate length and direction) from
the end of vector 𝑨.
3. The resultant 𝑹 is drawn from the
origin of 𝑨 to the end of 𝑩.
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= ?
! !
A+ B
!
B
!
A
8
Parallelogram Method
!
A
+
!
B
=
! !
A+ B = ?
Solution:
1. Draw the first vector 𝑨 (with the appropriate
length and direction).
2. Draw the next vector 𝑩 (with the appropriate
length and direction) from the origin of vector 𝑨.
3. From the segment of vectors 𝑨 and 𝑩, draw a
parallelogram (4 sided-polygon with parallel
opposite sides of equal length.
4. The resultant 𝑹 is drawn as a diagonal from the
origin of 𝑨 and 𝑩.
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!
B
! !
A+ B
!
A
9
Polygon Method
q In adding more than 2 vectors, use the Tail-to-tip
method or Triangle method.
+
𝑨
+
𝑩
𝑪 +
= 𝑨+𝑩+𝑪+𝑫 = 𝑹= ?
𝑫
𝑫
𝑫
𝑹
𝑪
𝑨
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𝑩
𝑹
𝑨
+
𝑩
+
𝑪
𝑨+𝑩
𝑨
𝑪
𝑩
10
Vector Subtraction
Special case of vector
Subtraction of Vectors.
addition:
Consider vectors 𝑨 and 𝑩. Recall the no. 3
property of vector addition:
𝑨+𝑩
𝑨
𝑨 − 𝑩 = 𝑨 + −𝑩
q 𝑨 − 𝑩 is just the addition of vector 𝑨
and −𝑩 (equal in magnitude, opposite in
direction of 𝑩)
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𝑩
𝑨−𝑩
−𝑩
11
Algebraic or Component
Method
Vectors: Described by the number, units and direction!
For example: Your displacement is 1.5 m at an angle of 250.
v Any vector 𝑨 (with a given magnitude and direction) in the xyplane can be represented as a VECTOR SUM of at least two
component vectors of 𝑨. As such, the components of 𝑨 in xyplane are 𝑨𝒙 and𝑨𝒚 .
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Algebraic or Component
Method
By components, your displacement is 1.36 m in the positive x
direction and 0.634 m in the positive y direction.
ü How to solve for the components of a vector?
à Use of trigonometric functions: sine and cosine functions
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14
Components of a Vector
Recall the definition of sine and cosine
function:
𝑦
𝒔𝒊𝒏𝒆 𝒐𝒇 𝒕𝒉𝒆 𝒂𝒏𝒈𝒍𝒆 =
𝒔𝒊𝒅𝒆 𝒐𝒑𝒑𝒐𝒔𝒊𝒕𝒆 𝒕𝒉𝒆 𝒂𝒏𝒈𝒍𝒆
𝒉𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆
and
𝑨
𝛉
𝒄𝒐𝒔𝒊𝒏𝒆 𝒐𝒇 𝒕𝒉𝒆 𝒂𝒏𝒈𝒍𝒆 =
𝑨𝒚 =?
Right angle
𝑨𝒙 =?
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𝒔𝒊𝒅𝒆 𝒂𝒅𝒋𝒂𝒄𝒆𝒏𝒕 𝒕𝒉𝒆 𝒂𝒏𝒈𝒍𝒆
𝒉𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆
v If the angle θ of vector 𝑨 is measured
from the x-axis:
𝑥 cos 𝜃 = !!! à 𝑨𝒙 = 𝑨 𝐜𝐨𝐬 𝜽
sin 𝜃 =
!"
!
à 𝑨𝒚 = 𝑨 𝐬𝐢𝐧 𝜽
(x-component of A)
(y-component of A)
15
Components of a Vector
𝑦
𝑨𝒙 =?
Recall the definition of sine and cosine
function:
𝒔𝒊𝒏𝒆 𝒐𝒇 𝒕𝒉𝒆 𝒂𝒏𝒈𝒍𝒆 =
Right angle
𝑨𝒚 =?
𝛉
𝒔𝒊𝒅𝒆 𝒐𝒑𝒑𝒐𝒔𝒊𝒕𝒆 𝒕𝒉𝒆 𝒂𝒏𝒈𝒍𝒆
𝒉𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆
and
𝑨
𝒄𝒐𝒔𝒊𝒏𝒆 𝒐𝒇 𝒕𝒉𝒆 𝒂𝒏𝒈𝒍𝒆 =
𝒔𝒊𝒅𝒆 𝒂𝒅𝒋𝒂𝒄𝒆𝒏𝒕 𝒕𝒉𝒆 𝒂𝒏𝒈𝒍𝒆
𝒉𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆
v If the angle θ of vector 𝑨 is measured
from the y-axis:
𝑥 sin 𝜃 = !! à 𝑨 = 𝑨 𝐬𝐢𝐧 𝜽
𝒙
!
cos 𝜃 =
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!"
!
à 𝑨𝒚 = 𝑨 𝐜𝐨𝐬 𝜽
(x-component of A)
(y-component of A)
16
Components of a Vector
Ø The components can be positive or negative (depending on
where the vector is located on the xy-plane) and will have
the same units as the original vector.
II
y
I
II
Ax < 0
Ay < 0
III
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Ax > 0
Ay < 0
IV
I
Ax = Acos90 = 0
Ay = Asin90 = A
Ax > 0
Ay > 0
Ax < 0
Ay > 0
y
x
Ax = Acos0 = A
Ay = Asin0 = 0
x
Ax = Acos180 = -A
Ay = Asin180 = 0
Ax = Acos270 = 0
Ay = Asin270 = -A
III
IV
17
Components of a Vector
q In case the vector 𝑨 is unknown but only its component
vectors are given, then we can solve for the magnitude and
the direction of the unknown vector.
𝑦
q Using Pythagorean Theorem, the
magnitude of 𝑨 is
𝐴& = 𝐴' & + 𝐴( & à 𝑨 =
𝑨 =?
𝑨𝒚
q Using the tangent function, the
direction of 𝑨 is
𝛉 =?
𝑨𝒙
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𝑨𝒙 𝟐 + 𝑨𝒚 𝟐
*3
𝑥
tan 𝜃 = * à 𝜽 =
4
𝑨𝒚
+𝟏
𝐭𝐚𝐧
𝑨𝒙
18
Components of a Vector
Ø Where should the θ be appropriately placed in the Cartesian
xy-plane?
N
II
θ°W of N
I
II
W
θ°N of E
θ°S of W
θ°S of E
θ°W of S
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W
45°N of W
45°N of E
45°S of W
45°S of E
45°W of S
θ°E of S
III
45°E of N
Northwest (NW)
E
I
Northeast (NE)
45°W of N
θ°E of N
θ°N of W
N
S
IV
Southwest (SW)
III
E
Southeast (SE)
45°E of S
S
IV
19
Unit Vectors
v Components of a vector are also vector
quantities:
y
2D: 𝑨 = 𝑨𝒙 + 𝑨𝒚
𝚥̂
k>
z
<i
v The direction of the components of 𝑨
are described by Unit Vectors i-hat, jhat, k-hat
x
ˆj ® y
kˆ ® z
iˆ ® x
v Unit vectors have a magnitude of 1
:
v 𝑨𝒙 = 𝑨𝒙 8;̂ 𝑨𝒚 = 𝑨𝒚 9;̂ and 𝑨𝒛 = 𝑨𝒛𝒌
v 𝑨 = 𝑨𝒙 8̂ + 𝑨𝒚 9̂
:
𝑨𝒛𝒌
Magnitude + Sign
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3D: 𝑨 = 𝑨𝒙 + 𝑨𝒚 + 𝑨𝒛
or 𝑨 = 𝑨𝒙 8̂ + 𝑨𝒚 9̂ +
Unit vector
20
Unit Vectors
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21
Dot (Scalar) Product
𝑨 " 𝑩 = 𝑨 𝑩 𝐜𝐨𝐬 𝜽
𝜽
𝑨#𝑩>𝟎
𝑨#𝑩<𝟎
𝑨#𝑩=𝟎
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𝜽
0° ≤ 𝜃 < 90°
90° < 𝜃 ≤ 180°
𝜃 = 90°
22
Property of Dot Product
1. Commutative
2. Distributive
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𝑨#𝑩=𝑩#𝑨
𝑨# 𝑩+𝑪 =𝑨#𝑩+𝑨#𝑪
23
Cross (Vector) Product
,
𝑨×𝑩 = 𝑨 𝑩 𝐬𝐢𝐧 𝜽 𝒏
§ A vector whose magnitude
equal to the product of A,B and
the sine of the angle between
A and B, and whose direction is
perpendicular to both A and B.
§ We can use unit vectors or
the right hand rule in
performing a cross product.
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Cross (Vector) Product
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25
Cross (Vector) Product
Vector Product
First Vector
Second Vector
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26
Problem 1.1
q
A particle is displaced by 5.25 m, 35° west of
south from its origin.
q
What are the components of its displacement?
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Problem 1.2
A man travels according to these series of location
𝐴⃗ = 1.75 𝑘𝑚, 150° 𝑐𝑤 𝑤𝑟𝑡 + 𝑥 − 𝑎𝑥𝑖𝑠
𝐵 = 2.00 𝑘𝑚, 15° 𝐸 𝑜𝑓 𝑁
𝐶⃗ = 2.25 𝑘𝑚, 𝑆𝑊
𝐷 = 3.0 𝑘𝑚, 𝑆
A. Draw his path of travel. (3 points)
B. Find his displacement from the origin. (7 points)
C. How far and what direction should he traverse if he
wants to have a final destination of
3.00 𝑘𝑚, 15° 𝑆 𝑜𝑓 𝐸 from his original location? (5
points)
30
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q
Problem 1.3
q
The vectors are shown in the figure. Find the following
(a) 𝑨 + 𝑩
(b)𝑨 − 𝑩
𝑦
𝑩 = 𝟐𝟎 𝒎
𝜽 = 𝟑𝟕°
𝑨 = 𝟕𝒎
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𝑥
31
Problem 1.4
q
Vector 𝐴⃗ is 2.80 cm long and is 60° above the x-axis in
the first quadrant.
q
Vector 𝐵 is 1.90 cm long and is below the x-axis in the
fourth quadrant.
q
Find the resultant vector 𝑅
of 𝐴⃗ and 𝐵.
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Problem 1.5
q
Three forces when added produce a resultant force
that is 100 N directed 63° above the +x axis.
q
If one of the forces is 50 N in the +y direction and
another 100 N in a direction 45° below the +x-axis,
what is the magnitude and the direction of the third
vector?
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Problem 1.6
q
Given with the following four vectors:
𝑨 = 20 𝑘𝑚, 𝑒𝑎𝑠𝑡
𝑩 = 56 𝑘𝑚, 𝑛𝑜𝑟𝑡ℎ𝑒𝑎𝑠𝑡
𝑪 = 11 𝑘𝑚, 22° 𝑠𝑜𝑢𝑡ℎ 𝑜𝑓 𝑒𝑎𝑠𝑡
𝑫 = 88 𝑘𝑚, 44° 𝑤𝑒𝑠𝑡𝑜𝑓 𝑠𝑜𝑢𝑡ℎ
A. Sketch the graph of the resultant vector using polygon (tail-totip) method. (2 points)
B. Find the x-component of the resultant vector. (3 points)
C. Find the y-component of the resultant vector. (3 points)
D. Calculate the magnitude and direction of the resultant vector
𝑅. (2 points)
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Problem 1.7
q
Josiah, Von, and Ely are standing in a strawberry field.
Von is 14.0 m due west of Josiah. Ely is 36.0 m from
Von, in a direction 37.0° south of east from Von’s
location.
A. How far is Ely from Josiah?
B. What is the direction of Ely’s location from that of
Josiah?
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Problem 1.8
A person walks 60 m at 47° north of east, turns and
walks 72 m at 15° south of east, and then turns and
walks 30 m, 30° west of north.
(a) How far and at what angle is the person's final position
from his/her initial position? (8 points)
q
(b) In what direction (displacement vector) would the
person have to head to return to his/her initial position?
(2 points)
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Problem 1.9
q
q
q
q
q
Kim and John decided to leave school and meet at the
plaza.
Kim directly walks 800 m east to reach the plaza.
John, on the other hand, first went to a store to buy
food which is 300 m southeast of school.
Then he walks to a flower shop to buy flowers which is
900 m, 60° E of N of the store.
What should be the displacement of John for him to
reach the plaza where Kim is waiting?
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Problem 1.10
q
q
q
An explorer in a dense jungles of equatorial Africa
leaves his hut.
He takes 40 steps northeast, then 80 steps 60° north
of west, then 50 steps due south.
Assume his steps all have equal length. Save him from
becoming hopelessly lost in the jungle by giving him
the displacement, calculated using the method of
components, that will return him to his hut.
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Seatwork
q
q
q
q
Glenvel and Sheena are standing under a tree in the
middle of a pasture.
An argument ensues, and they walk away in different
directions. Glen walks 2.0 ft/s in a direction 60.0° west
of north.
Sheen walks 1.5 ft/s in a direction 30.0° south of west.
They then stop and turn to face each other.
What is the distance and direction should Glen walk to
go directly toward Sheen?
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Problem 1.11
q Find the scalar product of the two vectors in
the figure. The magnitudes of the vectors are
𝐴 = 4.00 and 𝐵 = 5.00.
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40
Problem 1.12
q Find the angle between vectors
𝐴 = 2.00𝚤̂ + 3.00𝚥̂ + 1.00𝑘? and
?
𝐵 = −4.00𝚤̂ + 2.00𝚥̂ − 1.00𝑘.
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41
Problem 1.13
q Find the scalar product of the two vectors in
the figure. The magnitudes of the vectors are
𝐴 = 4.00 and 𝐵 = 5.00.
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42
Problem 1.14
qYou are given vectors 𝐴⃗ = 5.0𝚤̂ − 6.5𝚥̂ and 𝐵 =
− 3.5𝚤̂ + 7 𝚥̂ .
qA third vector lies in the xy-plane. Vector 𝐶⃗ is
perpendicular to vector 𝐴⃗ and the scalar
product of 𝐶⃗ with 𝐵 is 15.0.
qFrom this information, find the components of
vector 𝐶⃗
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Problem 1.15
(a) Find 𝑨 # 𝑩 for
?
𝐴⃗ = 2𝚤̂ + 3𝚥̂ – 𝑘? and 𝐵 = −1𝚤̂ + 3 𝚥̂ – 5𝑘.
⃗
(b) Let 𝐶⃗ = 𝐴⃗ − 𝐵. Calculate 𝐶⃗ # 𝐶.
(c) Find 𝑨×𝑩 for
?
𝐴⃗ = 2𝚤̂ + 3𝚥̂ – 𝑘? and 𝐵 = −1𝚤̂ + 3 𝚥̂ – 5𝑘.
⃗ 𝐵, and 𝐶,
⃗
(d) Prove that for any three vectors 𝐴,
𝑨 # 𝑩×𝑪 = 𝑨×𝑩 # 𝑪
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Problem 1.16
qThree vectors are given as follows:
?
𝐴⃗ = −2𝚤̂ – 5𝚥̂ + 2𝑘,
𝐵 = −5𝚤̂ − 2𝚥̂ − 3𝑘? 𝑎𝑛𝑑
?
𝐶⃗ = 4𝚤̂ + 1𝚥̂ + 6𝑘.
qCalculate the scalar product 𝐵 # 𝐴⃗ (2 points)
qCalculate the vector product 𝐵×𝐶⃗ (2 points)
⃗ (2
qCalculate the triple product of 𝐴⃗ # (𝐵×𝐶).
points)
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45
enables us to describe motion.
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46
Motion
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change in position of an object
with respect to time.
47
Rectilinear Motion
ü Motion of an object along a straight line.
ü Simplest kind of motion: One (1) dimensional motion.
ü A component of a more complex motion: two (2) or three
dimensional motion.
4 Basic Quantities of Kinematics
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48
Displacement
o is a change of position over a
time interval.
o Displacement: ∆𝒙
∆𝒙 = 𝒙 − 𝒙𝟎
o No subscript stands for final
and 0 stands for initial.
o It has both magnitude and
direction: + or – sign.
o SI Unit: meters (m).
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x (m)
0
1
2
3
4
5
6
x1 = + 1.0 m
x2 = + 7.2 m
Δx = + 7.2 m - 1.0 m = +6.2 m
7
x1 = + 2.5 m
x2 = - 2.0 m
Δx = -2.0 m - 2.5 m = -4.5 m
x1 = - 3.0 m
x2 = + 1.0 m
Δx = +1.0 m + 3.0 m = +4.0 m
50
Speed and Velocity
𝐬𝐩𝐞𝐞𝐝: 𝒗𝐚𝐯𝐞
𝒕𝒐𝒕𝒂𝒍 𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝐝
=
=
time interval
∆𝐭
(𝑠𝑐𝑎𝑙𝑎𝑟 𝑞𝑢𝑎𝑛𝑡𝑖𝑡𝑦)
𝐯𝐞𝐥𝐨𝐜𝐢𝐭𝐲: 𝒗𝒂𝒗𝒆
Instantaneous Speed/Velocity
– velocity at a specific instant of time
or specific point along the path.
q “instant” [physics] – refers to a
𝒅𝒊𝒔𝒑𝒍𝒂𝒄𝒆𝒎𝒆𝒏𝒕
single value of time, it has no
=
duration at all.
time interval
(𝑣𝑒𝑐𝑡𝑜𝑟 𝑞𝑢𝑎𝑛𝑡𝑖𝑡𝑦)
𝒙𝟐 − 𝒙𝟏
=
𝒕𝟐 − 𝒕𝟏
𝒗𝒂𝒗𝒆
Ø Average velocity tells the velocity of
the path travelled for the whole
duration.
∆𝐱
=
∆𝐭
üSI Unit: meters per second [m/s].
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52
Rules in Determining the Sign
for Velocity
Initial and Final Positions
Initial (+ or –)
Final (+ or –) < Initial
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Motion
Sign of
Velocity
Object is
moving the +
direction
Positive
Object is
moving the
– direction
Negative
Final (+ or –) > Initial
Initial (+ or –)
54
Acceleration
q The rate at which the velocity changes during time
interval.
𝒂𝒂𝒗𝒆
𝒗 − 𝒗𝟎 ∆𝒗
=
=
∆𝒕
∆𝒕
q Three situations in which an object is accelerating
(considering the velocity):
(a) Magnitude is changing, direction is constant
(b) Magnitude is constant, direction is changing
(c) Both magnitude and direction are changing.
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55
Rules in Determining the
Sign for Acceleration
Initial and Final Velocities
Initial:
Final: (more +)
𝒗𝒇 > 𝒗𝒊
𝒗𝒊 ≥ 𝟎 (+)
Motion
Sign of
Acceleration
The
motion of
object
speeds up
Positive
Object moves in + direction
Initial:
𝒗𝒊 > 𝟎 (+)
Final: (less +)
𝟎 ≤ 𝒗𝒇 < 𝒗𝒊
Object moves in + direction
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The
motion of
Negative
object
slows
down
56
Rules in Determining the
Sign for Acceleration
Motion
Initial and Final Velocities
Sign of
Acceleration
Initial:
Final: (more –)
𝒗𝒇 < 𝒗𝒊
𝒗 𝒊 ≤ 𝟎 (– )
The
motion of Negative
object
speeds up
Object moves in – direction
Initial:
Final: (less –)
𝟎 ≥ 𝒗𝒇 > 𝒗𝒊
Object moves in – direction
9/8/22
The
𝒗𝒊 < 𝟎 (–) motion of
object
slows
down
Positive
57
Motion with Constant
Acceleration
q The motion of a body undergoing a constant acceleration is
defined by the following equations:
Equation 1:
Equation 2:
Equation 3:
Equation 4:
𝑣' = 𝑣.' + 𝑎' 𝑡
1
𝑥 − 𝑥. = 𝑣.' 𝑡 + 𝑎' 𝑡 &
2
1
𝑥 − 𝑥. = 𝑣.' + 𝑣' 𝑡
2
𝑣𝑥 &−𝑣.' &= 2𝑎' (𝑥 − 𝑥0)
à 4 Kinematic Equations
Note: For a uniform (constant) acceleration, the instantaneous
accelerations are equal to the average acceleration.
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58
Problem 1.16
q A man walks 25 m due north. He turns and continues
walking 30 m due east.
q Calculate the distance travelled and the displacement
of the man from its starting point.
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59
Problem 1.17
q A group of five students decided to have an
educational tour in the industrial plants of the city.
q The tour guide begins to drive at 8:00 am and stops at
4:00 pm.
q Within this period, the car covers 260 km and 180 km
due north of its starting point.
q Find the speed and average velocity for the day.
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60
Problem 1.18
q A bicycle accelerates to a velocity 0.5 m/s in 5.0s.
The average acceleration over this time interval is
0.5 m/s2.
q What was the initial velocity of the bike?
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61
Problem 1.19
q A car’s speedometer reads 35.0 km/hr at one instant.
q Running along the same direction, after 25.0 s, it reads
40 km/hr.
q Calculate the magnitude of the average acceleration of
the car?
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62
Problem 1.20
q A ball rolls down a ramp fro 15 seconds.
q If the initial velocity of the ball was 0.8 m/s and the
final velocity was 7 m/sec, what was the acceleration
of the ball?
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63
Problem 1.21
q An airplane accelerates down a runway at 3.20
m/s2 for 32.8 s until is finally lifts off the ground.
q Determine the distance travelled before take off.
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64
Problem 1.22
q A car starts from rest heading due east. It first
accelerates at 3.00 m/s2 for 5.0 s and then continuous
without further acceleration for 20.0 s. It then brakes
for 8.0 s in coming to rest.
(a) What is the car’s velocity after the first 5.0 sec?
(b) What is the car’s acceleration over the last 8.0 s
interval?
(c) What is the total displacement?
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65
Problem 1.23
q A bus moving along a straight road at 25 m/s is
increasing its speed at the rate of 1.0 m/s each
second.
(a) Find the distance covered in 5s.
(b) If its speed is decreasing at the rate of 2 m/s, find the
distance traversed in 6 sec and
(c) The time it takes the bus to come to rest.
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66
Problem 1.24
q A car starts from rest and accelerates in a straight
line at 1.6 m/s2 for 10 seconds.
(a) What is its final speed?
(b) How far has it travelled in this time?
(c) If the brakes are then applied and it travels a
further 20 m before stopping, what is the
deceleration?
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67
Problem 1.25
qA subway train starts from rest at a station and
accelerates at a rate of 𝟏. 𝟔𝟎 𝒎/𝒔𝟐 for 𝟏𝟒. 𝟎 𝒔.
qIt runs at a constant speed for 𝟕𝟎. 𝟎 𝒔 and slows
down at a rate of 𝟑. 𝟓𝟎 𝒎/𝒔𝟐 until it stops at the
next station.
qFind the total distance
covered.
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68
Problem 1.26
• A ball starts from rest and rolls down a hill with
uniform acceleration, traveling 150 m during the
second 5.0 s of its motion.
A. Find the expression of the final velocity for the
first 5.0 s. (2 points)
B. Calculate for the magnitude of the uniform
acceleration. (4 points)
C. How far did it roll during the first 5.0 s of
motion? (4 points)
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69
Problem 1.27
q At the instant the traffic light turns green, a car that
has been waiting at an intersection starts ahead
with a constant acceleration of 3.20 𝑚/𝑠 \ .
q At the same instant a truck, traveling with a
constant speed of 20.0 𝑚/𝑠 overtakes and passes
the car.
(a) How far beyond its starting point does the car
overtake the truck?
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70
Problem 1.28
• A car and a bus are traveling along the same straight
road in neighbouring lanes.
• The car has a constant velocity of +25.0 𝑚/𝑠, and at
𝑡 = 0 it is located 21.0 𝑚 ahead of the bus.
• At time 𝑡 = 0, the bus has a velocity of +5.0 𝑚/𝑠 and
an acceleration of +2.0 𝑚/𝑠 K .
(a) When does the bus pass the car?
(b) How far beyond its starting point does the bus
overtake the car?
(c) What is the velocity of the bus at this point (in letter b)?
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Problem 1.29
• An object’s velocity is measured to be
𝒗𝒙 𝒕 = 𝜶 − 𝜷𝒕𝟐 ,
where 𝛼 = 4.00 𝑚/𝑠 and 𝛽 = 2.00 𝑚/𝑠 L . At 𝑡 = 0 the
object is at 𝑥 = 0.
(a) Calculate the object’s position and acceleration as a
function of time.
(b) What is the object’s maximum positive displacement
from the origin?
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72
Free Fall Motion
Ø Motion of any body or object that is under the influence of the
earth’s gravity.
Ø Most familiar case of motion with constant acceleration, 𝒂 =
−𝒈
𝒈 = 𝟗. 𝟖 𝒎⁄𝒔𝟐 (acceleration due to gravity)
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73
Free Fall Motion
Q: If you dropped two objects of different
masses from a certain height from the
ground at the same time, which object
would arrive/hit the ground first?
The heavier object?
The lighter object?
A: Both objects would arrive on the ground at
the same time.
ü Ideally, free fall motion is independent on
the mass of the object.
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74
Free Fall Motion
Ideally, free fall motion is independent on
the mass of the object. Only if:
o Air resistance (effects
ignored/neglected.
of
air)
is
o The distance of the fall is small compared
to the radius of the earth; and
o Effect of the earth’s rotation is ignored.
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75
Free Fall Motion Equations
q Since free fall motion is a case of motion with constant acceleration
(𝒂𝒚 = −𝒈), then the 4 Kinematic Equations also apply to free fall.
Equation 1:
𝑣' = 𝑣.' + 𝑎𝑡
𝑣𝑦 = 𝑣.( − 𝑔𝑡
1 &
Equation 2: 𝑥 − 𝑥. = 𝑣.' 𝑡 + 𝑎𝑡
2
1
𝑣.' + 𝑣' 𝑡
Equation 3: 𝑥 − 𝑥. =
2
1 &
𝑦 = 𝑦. + 𝑣.( 𝑡 − 𝑔𝑡
2
& − 𝑣 & = 2𝑎(𝑥 − 𝑥 )
𝑣
'
.'
0
Equation 4:
𝑣𝑦2 = 𝑣.( & −2𝑔 (𝑦 − 𝑦.)
1
𝑦 = 𝑦. + 𝑣.( + 𝑣𝑦 𝑡
2
where acceleration due to gravity 𝒈 = 𝟗. 𝟖 𝒎⁄𝒔𝟐
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76
Problem 1.30
q A steel ball dropped from a tower strikes a
ground in 4.0 s. Find:
(a) the velocity 𝒗𝒚 with which the ball strikes
the ground, and
(b) the height of the tower.
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77
Problem 1.31
qA
large boulder is ejected
vertically upward from a volcano
with an initial speed of 40.0 m/s.
Air resistance may be ignored.
(a) At what time after being ejected is the boulder moving
at 20.0 m/s upward?
(b) At what time is it moving at 20 m/s downward?
(c) When is the displacement of the boulder from its initial
position zero?
(d) When is the velocity of the boulder zero?
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78
Problem 1.32
q A student throws a water balloon vertically
downward from the top of a building.
q The balloon leaves the thrower's hand
with a speed of 6.00 m/s .
q Air resistance may be ignored, so the
water balloon is in free fall after it leaves
the thrower's hand.
(a) What is its speed after falling for a time
2.00 s ?
(b) How far does it fall in 2.00 s ?
(c) What is the magnitude of the velocity after
falling 10.0 m?
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79
Problem 1.33
A hot-air balloonist, rising
vertically with a constant velocity
of magnitude 5.00 m/s, releases a
sandbag at an instant when the
balloon is 40.0 m above the
ground.
q After it is released, the sandbag is
in free fall.
(a) Compute the position and velocity
of the sandbag at 0.250 s and 1.00 s
after its release.
q
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80
Problem 1.34
(b) How many seconds after its release will the bag strike
the ground?
(c) With what magnitude of velocity does it strike the
ground?
(d) What is the greatest height above the ground that the
sandbag reaches?
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81
Problem 1.35
q
q
You are on the roof of the physics building, 46.0 m
above the ground. Your physics professor, who is 1.80
m tall, is walking alongside the building at a constant
speed of 1.20 m/s.
If you wish to drop an
egg on your professor’s
head, where should the
professor be when you
release the egg? Assume
that the egg is in free fall.
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82
Problem 1.36
q
A certain freely falling object requires 1.50 s to
travel the last 30.0 m before it hits the ground.
From what height above the ground did it fall?
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83
Problem 1.37
Sam heaves a 16-lb shot straight upward, giving it a
constant upward acceleration from rest of for 64.0 cm.
q He releases it 2.20 m above the ground. You may
ignore air resistance.
(a) What is the speed of the shot when Sam releases it?
(b) How high above the ground does it go?
(c) How much time does he have to get out of its way
before it returns to the height of the top of his head, 1.83
m above the ground?
q
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84
Kinematic Equations (1D)
q The motion of a body undergoing a constant acceleration is
defined by the following equations:
x-component
y-component
Equation 1:
𝑣𝑦 = 𝑣.( − 𝑔𝑡
𝑣' = 𝑣.' + 𝑎𝑡
1 &
Equation 2:
𝑦 − 𝑦. = 𝑣.( 𝑡 − 𝑔𝑡
1 &
2
𝑥 − 𝑥. = 𝑣.' 𝑡 + 𝑎𝑡
2
1
Equation 3: 1
𝑦 − 𝑦. = 𝑣.( + 𝑣𝑦 𝑡
𝑥 − 𝑥. = 𝑣.' + 𝑣' 𝑡
2
2
Equation 4:
2 = 𝑣 & −2𝑔 (𝑦 − 𝑦 )
&
&
𝑣
𝑦
.(
.
𝑣' − 𝑣.' = 2𝑎(𝑥 − 𝑥0)
à 4 Kinematic Equations
Note: For a uniform (constant) acceleration, the instantaneous
accelerations are equal to the average acceleration.
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Projectile Motion
Trajectory - the path followed by a projectile. (Parabola)
Characteristic of a Projectile Motion:
i. Constant vertical acceleration (free-fall motion)
ii. Constant horizontal motion (constant speed, zero acceleration)
where
𝒗𝟎𝒙 = 𝒗𝟎 cos 𝜽(
𝒗𝟎𝒚 = 𝒗𝟎 sin 𝜽(
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86
Equations of Motions of a
Projectile
Horizontal Motion
ü Since the velocity is constant, then
the horizontal acceleration 𝒂𝒙 = 𝟎
Eqn 1: 𝒗𝒙 = 𝒗𝟎𝒙 + 𝒂𝒙𝒕
Eqn 4: 𝒗𝒙𝟐 = 𝒗𝟎𝒙𝟐 + 𝟐𝒂𝒙 (𝒙 − 𝒙𝟎)
𝒗𝒙 = 𝒗𝟎𝒙
𝟏
Eqn 2: 𝒙 = 𝒙𝟎 +𝒗𝟎𝒙 𝒕 + 𝟐 𝒂𝒙
Eqn 3: 𝒙 =
𝟏
𝟐
𝒕𝟐
𝒗𝒙 + 𝒗𝟎𝒙 𝒕
Vertical Motion
ü The vertical motion of the projectile
has a constant acceleration
𝒂𝒚 = −𝒈;
𝒈 = 𝟗. 𝟖 𝐦⁄𝐬 𝟐
Eqn 1: 𝒗𝒚 = 𝒗𝟎𝒚 − 𝒈𝒕
𝟏
Eqn 2: 𝒚 = 𝒚𝟎 + 𝒗𝟎𝒚 𝒕 − 𝒈𝒕𝟐
Eqn 3: 𝒚 = 𝒚𝟎 +
𝟏
𝟐
𝟐
𝒗𝒚 + 𝒗𝟎𝒚 𝒕
Eqn 4: 𝒗𝒚 𝟐 = 𝒗𝟎𝒚 𝟐 − 𝟐𝒈(𝒚 − 𝒚𝟎)
𝒙 = 𝒙𝟎 + 𝒗𝟎𝒙 𝒕
where
9/8/22
𝒗𝟎𝒙 = 𝒗𝟎 cos 𝜽(
𝒗𝟎𝒚 = 𝒗𝟎 sin 𝜽(
87
Equations of Motions of a
Projectile
o At any time 𝒕, the position of the projectile is given by 𝒙 = 𝒗𝟎𝒙 𝒕 and 𝒚 = 𝒗𝟎𝒚 𝒕 +
𝟏
− 𝒈𝒕𝟐
𝟐
Then 𝒕 =
𝒙
𝒗𝟎𝒙
=
𝒙
,
𝒗𝟎 𝐜𝐨𝐬 𝜽
𝟏
𝟐
substituting to 𝒚 = 𝒗𝟎𝒚 𝒕 − 𝒈𝒕𝟐 .
1 '
𝑦 = 𝑣%& 𝑡 − 𝑔𝑡
2
𝑥
1
𝑥
𝑦 = 𝑣% sin 𝜃%
− 𝑔
𝑣% cos 𝜃%
2 𝑣% cos 𝜃%
sin 𝜃%
1
𝑥'
𝑦=𝑥
− 𝑔 '
cos 𝜃%
2 𝑣% cos ' 𝜃%
𝒚 = 𝒙 𝐭𝐚𝐧 𝜽𝟐 −
𝒈
𝟐𝒗𝟎 𝟐 𝐜𝐨𝐬 𝟐 𝜽𝟎
'
𝒙𝟐
(equation of a parabola)
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Maximum Height of a
Projectile
o At the maximum height 𝒉𝒎𝒂𝒙 = 𝒚 , the vertical velocity is zero (𝒗𝒚 = 𝟎) but
horizontal velocity remains constant (𝒗𝒙 = 𝒗𝟎𝒙 = 𝒗𝟎 𝐜𝐨𝐬 𝜽𝟎 ). From Eqn. 4:
0 = 𝑣(- . + 2𝑎- ℎ/01
2𝑎- ℎ/01 = −𝑣(- .
2𝑎- ℎ/01 = −𝑣( . sin. 𝜃(
𝒗𝟎 𝟐 𝐬𝐢𝐧𝟐 𝜽𝟎
𝒉𝒎𝒂𝒙 = −
𝟐𝒂𝒚
o Using Eqn. 1, the time needed for the
projectile to reach the maximum
height is:
0 = 𝑣(- + 𝑎- 𝑡
𝒉𝒎𝒂𝒙
𝑎- 𝑡 = −𝑣(- = −𝑣( sin 𝜃(
𝒗𝟎 𝐬𝐢𝐧 𝜽𝟎
𝒕=−
𝒂𝒚
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Maximum Range of a
Projectile
o The time when the vertical position of
projectile is equal to its initial vertical position
(𝑦 = 𝑦% = 0),
1
0 = 0 + 𝑣#$ 𝑡 + 𝑎& 𝑡 %
2
1
𝑎& 𝑡 % = −𝑣#$ 𝑡
2
𝟐𝒗𝟎 𝒔𝒊𝒏 𝜽𝟎
𝒕=−
𝒂𝒚
o At that instant, the projectile has covered
a maximum horizontal distance (range 𝑅),
𝑥 = 𝑣%) 𝑡
−2𝑣% sin 𝜃%
𝑅 = 𝑣% cos 𝜃%
𝑎&
−2𝑣% ' cos 𝜃% sin 𝜃%
=
𝑎&
Recall: sin 2𝛼 = sin 𝛼 cos 𝛼
𝒗𝟎 𝟐 𝐬𝐢𝐧 𝟐𝜽𝟎
𝑹=−
𝒂𝒚
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𝑹
o The velocity of the projectile at this
position is:
𝑣& = 𝑣%& + 𝑎& 𝑡
−2𝑣%&
𝑣& = 𝑣%& + 𝑎&
𝑎&
𝑣& = 𝑣%& − 2𝑣%&
𝒗𝒚 = −𝒗𝟎𝒚
90
Projectile Motion at
Various Initial Angles
A projectile launched at different angles with the same initial velocity
Ø
The heights will be different at
different projection angle (height
increases as the angle increases)
𝒉𝒎𝒂𝒙
Ø
𝒗𝟎 𝟐 𝐬𝐢𝐧𝟐 𝜽𝟎
=−
𝟐𝒂𝒚
The time of flight also increases
with the angle of projection
𝒕=−
Ø
𝟐𝒗𝟎 𝐬𝐢𝐧 𝜽𝟎
𝒂𝒚
The maximum range R occurs at a
projection angle of 45° (between
0° to 90°)
𝒗𝟎 𝟐 𝐬𝐢𝐧 𝟐𝜽𝟎
𝑹=−
𝒂𝒚
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Problem 1.38
q
The projectile fired with an initial speed of 25.0 m/s
oriented at 37° above the horizontal
(a) Find the position and the velocity of the ball when
𝑡 = 2.00 𝑠.
(b) Find the time when the ball reaches the highest point
of its flight and height h at this point.
(c) Find the total time of flight of the ball.
(d) Find the horizontal range R.
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Problem 1.39
q
A military helicopter on
training mission is flying Y0=300 m
horizontally at a speed of
60.0 m/s and accidentally
drops a bomb at an elevation
of 300 m.
q
Ignore air resistance.
(a) How much time is required for the bomb to reach the earth?
(b) How far does it travel horizontally while falling?
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93
Problem 1.39
(c) Find the horizontal and vertical components of its velocity
just before it strikes the earth?
(d) If the velocity of the helicopter remains constant, where is
the helicopter when the bomb hits the ground?
Y0=300 m
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Problem 1.40
q
q
A tennis ball rolls off the edge of a table top 0.750 m above
the floor and strikes the floor at a point 1.40 m horizontally
from the edge of the table.
Ignore air resistance.
(a) Find the time of flight.
(b) Find the magnitude of the
initial velocity.
(c) Find the magnitude and
direction of the velocity of the ball
just before it strikes the floor.
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Problem 1.41
q
q
John Elway throws a football with an initial upward velocity
component of 16.0 m/s and a horizontal velocity component of
20.0 m/s.
Ignore air resistance.
(a) How much time is required for the football to reach the highest
point of trajectory?
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96
Problem 1.42
(b) How high is this point?
(c) How much time (after being thrown) is required for the football
to return to its original level?
(d) How far has it travelled horizontally during this time?
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Problem 1.43
q
q
A shot putter releases the shot (heavy metal ball) some
distance above the level ground with a velocity of 12.0 𝑚/𝑠 at
an angle of 51.0° above the horizontal.
The shot hits the ground 2.08 s later.
(a) What are the components
of the shot’s acceleration?
(b) How far did he throw the
shot horizontally?
(c) How high was the shot
when it was released?
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Problem 1.44
A baseball thrown at an angle of 60° above the
horizontal strikes a building 18.0 m away at a point 8.00
m above the point from which it is thrown.
q Ignore air resistance.
q
(a) Find the magnitude of the initial velocity of the
baseball?
(b) Find the magnitude and direction of the velocity of
the baseball just before it strikes the building?
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Problem 1.45
A snowball rolls off a barn roof that slopes at an angle of
40° with respect to the horizontal.
q The edge of the roof is 14.0 m above the ground, and the
snowball has a speed of 7.0 m/s as it rolls off the roof.
q Ignore air resistance.
q
(a) How far from the edge of the barn does
the snowball strike the ground if it
doesn’t strike anything else while falling?
(b) A man 1.90 m tall is standing 4.0 m from
the edge of the barn. Will he be hit by
the snow ball?
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100
Problem 1.46
q
A projectile that fires a signal flare gives the flare an
initial speed of 120 m/s.
(a) If the flare is fired at an angle 55° above the horizontal
on the level salt flat of Utah, what is its horizontal range?
Ignore air resistance.
(b) If the flare is fired at the same
angle over a flat sea of tranquility
on the moon, where 𝑔 = 1.6 𝑚/𝑠2,
what is the horizontal range?
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101
Problem 1.47
q
In the given picture you see the motion path of
cannonball. Find the (a) maximum height it can reach,
(b) horizontal distance it covers and (c) total time
from the given information. (The angle between
cannonball and horizontal is 60°.
9/8/22
102
Problem 1.48
q
You are to throw a ball with a speed of 12.0 m/s at a target
that is height h = 5.00 m above the level at which you release
the ball. You want the ball’s velocity to be horizontal at the
instant it reaches the target.
(a) At what angle θ above the
horizontal must you throw the ball?
(b) How long does the ball hit the
target?
(c) What is the horizontal distance
from the release point to the target?
(d) What is the speed of the ball just
as it reaches the target?
9/8/22
103
Problem 1.49
Vincent did daredevil stunts in his spare time. His last stunt was
an attempt to jump across a river on a motorcycle as shown in
the right. The takeoff ramp was inclined at 53.0°, the river was
40.0 m wide, and the far bank was 15.0 m lower than the top of
the ramp. The river itself was
100 m below the ramp.
You can ignore air resistance.
(a) What should his speed have been at
the top of the ramp to have just made
it to the edge of the far bank? (17 m/s)
(b) If his speed was only half the value
found in part (a), where did he land?
(28.4 m)
q
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104