Chapter 3
Kinematics in Two
Dimensions; Vectors
• Vectors and Scalars
• Addition of Vectors – Graphical Methods (One and TwoDimension)
• Multiplication of a Vector by a Scalar
• Subtraction of Vectors – Graphical Methods
• Adding Vectors by Components
• Projectile Motion
• Projectile Motion Is Parabolic
• Relative Velocity
Adding Vectors by Components
Any vector can be expressed as the sum of two other vectors, which are called its
components. Usually the other vectors are chosen so that they are perpendicular
to each other.
Vx
Vy
In the figure above, the vector
points at an angle of 300 with respect to the +x
direction (in fact, it can point at any direction θ).
The component
is called the projection of
on the x axis with magnitude Vx
given by the line 0C, while the component
is the projection of
on the y axis
with magnitude Vy given by the line 0B.
Using the parallelogram method for vector addition:
(3.4)
Adding Vectors by Components
From now on, we will denote the magnitude of a vector by the following expression:
(3.5)
(3.6)
Where the symbol | | represents the module of the vector
magnitude.
We can also write
Note that the eq. 3.6 follows
from the Pythagorean
theorem.
, or, in other words, its
, and
Vx
Vy
Vy
Vx
Adding Vectors by Components
Example 3.4: Using the figure below, show that if
and
are two vectors in a
plane defined by two perpendicular axes x and y, and
is the resultant of the
addition of these two vectors, then:
y
(3.7)
x
Solution developed on the blackboard
Equation 3.7 states that the resultant of the sum of two vectors in a two dimensional
plane is equivalent to the sum of their components on each of the x and y axes.
Adding Vectors by Components
Equation 3.7 also applies when both vectors start away from the origin of the
coordinate system:
Adding Vectors by Components
If the components are perpendicular, they can be found using trigonometric functions.
In the figure, the vector makes and angle
θ with the x axis.
Note: We define (by convention) θ with
respect to the +x and +y (first quadrant)
directions:
θ
-θ
(3.8)
sin θ
= sine of theta
=
(3.9)
cos θ
= cosine of theta =
tan θ
= tangent of theta =
(3.10)
Adding Vectors by Components
You can use your calculator to compute sin, cos and tan. If you want to find the angle,
you can use the functions sin-1 , cos-1 and
tan-1 for equations 3.8, 3.9 and 3.10,
respectively. We write:
If you calculator does not have
these functions, it should have a key called
“invert” or “inv” or something similar.
Trigonometric functions can be added,
multiplied, etc when needed. You can find
several functions and identities in
Appendix A of the text book.
Three very important trigonometric identities
are the following:
(3.8)
(3.9)
(3.10)
Adding Vectors by Components
Using equations 3.8 – 3.10, we can write the components of the vector
as:
(3.11)
(3.12)
(3.8)
(3.9)
(3.10)
Adding Vectors by Components
Problem 3.10 (textbook): Three vectors are shown in Fig. 3–32. Their magnitudes are
given in arbitrary units. Determine the sum of the three vectors. Give the resultant in terms of
(a) components, (b) magnitude and angle with the x axis.
Fig. 3-32
10. Three vectors are shown in Fig. 3–32. Their magnitudes are given in arbitrary units.
Determine the sum of the three vectors. Give the resultant in terms of (a) components, (b)
magnitude and angle with the x axis.
Ax = 44.0 cos 28.0 o = 38.85
Ay = 44.0 sin 28.0 o = 20.66
Bx = −26.5 cos 56.0 o = −14.82
B y = 26.5 sin 56.0 o = 21.97
C x = 31.0 cos 270 o = 0.0
C y = 31.0 sin 270 o = −31.0
(a)
r r r
( A + B + C ) x = 38.85 + ( −14.82 ) + 0.0 = 24.03 = 24.0
r r r
( A + B + C ) y = 20.66 + 21.97 + ( −31.0 ) = 11.63 = 11.6
(b)
r r r
A+B+C =
θ = tan
−1
11.63
24.03
( 24.03 )
2
= 25.8 o
+ (11.63 ) = 26.7
2
Adding Vectors by Components
Problem 3.13 (textbook): For the vectors given in Fig. 3–32, determine
(b)
r r r
A − B + C,
r r r
A + B − C,
(c)
r r r
C − A − B.
(a)
Fig. 3-32
r r r
r r r
13. For the vectors given in Fig. 3–32, determine (a) A − B + C, (b) A + B − C, and
(c)
r r r
C − A − B.
(a)
Ax = 44.0 cos 28.0 o = 38.85
Ay = 44.0 sin 28.0 o = 20.66
Bx = −26.5 cos 56.0 o = −14.82
B y = 26.5 sin 56.0 o = 21.97
C x = 31.0 cos 270 o = 0.0
C y = 31.0 sin 270 o = −31.0
r r r
A
( − B + C ) x = 38.85 − ( −14.82 ) + 0.0 = 53.67
r r r
( A − B + C ) y = 20.66 − 21.97 + ( −31.0 ) = −32.31
Note that since the x component is positive and the y
component is negative, the vector is in the 4th quadrant.
r r r
A−B+C =
( 53.67 ) + ( −32.31) = 62.6
2
2
θ = tan −1
−32.31
53.67
= 31.0 o below + x axis
(b)
r r r
( A + B − C ) x = 38.85 + ( −14.82 ) − 0.0 = 24.03
r r r
( A + B − C ) y = 20.66 + 21.97 − ( −31.0 ) = 73.63
r r r
A+B−C =
( 24.03)
2
+ ( 73.63 ) = 77.5
2
θ = tan −1
73.63
24.03
= 71.9 o
(c)
r r r
( C − A − B ) x = 0.0 − 38.85 − ( −14.82 ) = −24.03
r r r
( C − A − B ) y = −31.0 − 20.66 − 21.97 = −73.63
Note that since both components are negative, the vector is in the 3rd quadrant.
r r r
C−A−B =
( −24.03 ) + ( −73.63 ) = 77.5
2
2
θ = tan −1
−73.63
−24.03
= 71.9 o below − x axis
Note that the answer to (c) is the exact opposite of the answer to (b).
Projectile Motion
Projectile motion is a generalization of purely vertical motion (falling objects) we
studied in chapter 2. Now we will consider the motion in two dimensions (can be
easily generalized to three dimensions).
An example of projectile motion is that of a tennis ball, speeding bullets, etc..
In this chapter, we will not consider how an object is set in motion, but only its motion
after it start moving and before it lands or is caught.
Projectile Motion
Projectile motions can be approached using the method of vector components as
described before.
We can analyze the motion by studying its vertical and horizontal components
separately (Galileo was the first to describe projectile motion using this method).
Example:
In this figure, a ball rolls off the end
of a table with an initial velocity Vx0
at an instant of time t = 0
in the horizontal direction.
Furthermore, its velocity
points in
the direction of the motion at any
instant of time t, and is tangent to its
path.
Projectile Motion
We can then analyze the motion in the x and y direction separately using the
equations of motion we obtained in chapter 2:
(2.10)
horizontal
motion
(2.13)
(2.14)
(2.16)
vertical
motion
(2.17)
(2.18)
Projectile Motion
y direction
We want to find the position and velocity (in the y-direction) at a given instant of time
“t”. In the vertical direction, we have the acceleration, g, of gravity. The initial velocity
is zero (Vy0 = 0). The coordinate y is chosen to be positive upward and zero at the
origin selected as where the point the ball start
experiencing vertical motion as depicted in the
figure. Then, at t = 0 we have:
y0 = 0
Vy0 = 0
a = –g
(“a” points downward)
Using equation 2.17 we have:
(3.13)
Using eq. 2.16:
(3.14)
Projectile Motion
x direction
In the horizontal direction, there is no acceleration. The initial velocity is Vx0. The
coordinate x is chosen to be eastward and zero at the origin selected as the point
where the ball start experiencing vertical motion as depicted in the figure.
Then, at t = 0 we have:
x0 = 0
initial velocity = Vx0
a = 0
Using equation 2.13 we have:
(3.15)
Using eq. 2.10:
(3.16)
The velocity is constant in the horizontal direction.
Projectile Motion
Example 3.5:
Show that an object projected horizontally will reach the ground in the same time as
an object dropped vertically from the same height (use the figure below).
The only motion that matters here is the vertical
(displacement in the y-direction to reach the ground).
Equation 3.13 gives the y displacement
for both balls since they are subject to similar
initial conditions, namely:
y0 = 0
Vy0 = 0
a = –g
(a points downward)
Therefore
applies in both cases at any instant of time.
Thus, the two ball reach the ground together.
Projectile Motion
If an object is launched at an initial angle of θ0 with the horizontal, the analysis is
similar except that the initial velocity has a vertical component.
You can use equations 3.11 and 3.12 and then develop the problem in the very same
way as on the previous slides.
Projectile Motion
Problem 3.21 (textbook) : A ball is thrown horizontally from the roof of a building
45.0 m tall and lands 24.0 m from the base. What was the ball’s initial speed?
Solution developed on the blackboard
Problem 3.21 (textbook) : A ball is thrown horizontally from the roof of a building
45.0 m tall and lands 24.0 m from the base. What was the ball’s initial speed?
Choose downward to be the positive y direction. The origin will be at the point where the ball is
thrown from the roof of the building. In the vertical direction,
vy 0 = 0
a y = 9.80 m s 2
.
y0 = 0
and the displacement is 45.0 m. The time of flight is found from applying Eq. 2.17 to the
vertical motion.
y = y0 + v y 0 t + 12 a y t 2
→
45.0 m =
1
2
(
)
9.80 m s 2 t 2
→
t=
2 ( 45.0 m )
9.80 m s
2
= 3.03 sec
The horizontal speed (which is the initial speed) is found from the horizontal motion at constant
velocity:
∆x = v x t
→ v x = ∆x t = 24.0 m 3.03 s = 7.92 m s
Projectile Motion
Problem 3.31 (textbook) : A projectile is shot from the edge of a cliff 125 m above
ground level with an initial speed of 65.0 m/s at an angle of 37.0º with the horizontal,
as shown in Fig. 3–35.
(a) Determine the time taken by the projectile to hit point P at ground level.
(b) Determine the range X of the projectile as measured from the base of the cliff.
At the instant just before the projectile hits point P:
(c) Find the horizontal and the vertical components of its velocity
(d) Find the magnitude of the velocity
(e) Find the angle made by the velocity vector with the horizontal.
(f ) Find the maximum height above the cliff top reached by the projectile.
Solution developed on the blackboard
Choose the origin to be at ground level, under the place where the projectile is launched, and
upwards to be the positive y direction. For the projectile,
v0 = 65.0 m s
θ 0 = 37.0o
ay = − g
y 0 = 125
v y 0 = v0 sin θ 0
(a)
The time taken to reach the ground is found from Eq. 2.17, with a final height of 0.
y = y0 + v y 0 t + 12 a y t 2
t=
→
0 = 125 + v0 sin θ 0 t − 12 gt 2
− v0 sin θ 0 ± v02 sin 2 θ 0 − 4 ( − 12 g ) (125 )
2(− g )
1
2
=
→
−39.1 ± 63.1
−9.8
= 10.4 s , − 2.45 s = 10.4 s
Choose the positive sign since the projectile was launched at time t = 0.
Choose the origin to be at ground level, under the place where the projectile is launched, and
upwards to be the positive y direction. For the projectile,
v0 = 65.0 m s
θ 0 = 37.0o
ay = − g
y 0 = 125
v y 0 = v0 sin θ 0
(b)
The horizontal range is found from the horizontal motion at constant velocity.
∆x = vx t = ( v0 cos θ 0 ) t = ( 65.0 m s ) cos 37.0 o (10.4 s ) = 541 m
Choose the origin to be at ground level, under the place where the projectile is launched, and
upwards to be the positive y direction. For the projectile,
v0 = 65.0 m s
θ 0 = 37.0o
ay = − g
y 0 = 125
v y 0 = v0 sin θ 0
(c)
At the instant just before the particle reaches the ground, the horizontal component of its
velocity is the constant
vx = v0 cos θ 0 = ( 65.0 m s ) cos 37.0o = 51.9 m s
The vertical component is found from Eq. 2.16:
(
v y = v y 0 + at = v0 sin θ 0 − gt = ( 65.0 m s ) sin 37.0 o − 9.80 m s 2
= −63.1m s
) (10.4 s )
Choose the origin to be at ground level, under the place where the projectile is launched, and
upwards to be the positive y direction. For the projectile,
v0 = 65.0 m s
θ 0 = 37.0o
ay = − g
y 0 = 125
v y 0 = v0 sin θ 0
(d)
The magnitude of the velocity is found from the x and y components calculated in part c) above.
v=
v x2 + v y2 =
( 51.9 m s )2 + ( −63.1m s )2
= 81.7 m s
Choose the origin to be at ground level, under the place where the projectile is launched, and
upwards to be the positive y direction. For the projectile,
v0 = 65.0 m s
θ 0 = 37.0o
ay = − g
y 0 = 125
v y 0 = v0 sin θ 0
(e)
The direction of the velocity is (see slide number 7 of this lecture)
θ = tan
−1
vy
vx
= tan
and so the object is moving
.
−1
− 63.1
51.9
= − 50.6 o
50.6o below the horizon
Choose the origin to be at ground level, under the place where the projectile is launched, and
upwards to be the positive y direction. For the projectile,
v0 = 65.0 m s
θ 0 = 37.0o
ay = − g
y 0 = 125
v y 0 = v0 sin θ 0
(f)
The maximum height above the cliff top reached by the projectile will occur when the yvelocity is 0, and is found from Eq. 2.18.
v y2 = v y2 0 + 2 a y
y m ax =
(y −
v 02 sin 2 θ 0
2g
=
y0 )
→
0 = v 02 sin 2 θ 0 − 2 g y m ax
( 6 5 .0 m s )2 sin 2 3 7 .0 o
(
2 9 .8 0 m s
2
)
= 7 8 .1 m
Assignment 1
Deadline changed to Tuesday, Sept. 29