Projectile Motion
Vectors –
Here we shall discuss motion in 2 dimensions
That is where the position of the object moves in a plane.
Scalar:
a quantity that possesses only magnitude
(i.e. the amount, how much).
e.g. mass, temperature, volume, time, speed etc.
These quantities obey the normal laws of addition and
subtraction.
Vector:
a quantity that possesses both magnitude and
direction (i.e. how much and in what direction).
e.g. velocity, acceleration, force etc.
Vector Representation
We represent a vector as a directed line segment with
the length of the line indicating the magnitude (or size) of
the quantity and the direction indicating where the quantity
is travelling. e.g. velocity of 5 m·s-1 East:
Note. Scale: 1 cm represents 1 m·s-1.
Conceptual Physics – 3rd Edition – Paul Hewitt
Chapter 3 – Projectile Motion
Page 1 of 9
In writing, a vector can be represented as a letter in
bold type (i.e. a) or as a letter with an arrow above it
or wriggly line below it (i.e. a or a)
Direction in Degrees True
If we state 0° T is North and other angles are
determined by the clockwise angle as we turn from
North, we are measuring direction by the convention
of degrees true. ( 0 T)
e.g.
0° T
90° T
Conceptual Physics – 3rd Edition – Paul Hewitt
180° T
Chapter 3 – Projectile Motion
270° T
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Triangle Law of Addition of Vectors
Draw a line with magnitude and direction to represent
the first mentioned vector AB.
• To the end of it draw a line with magnitude and
direction to represent the second mentioned vector BC.
• The resultant is obtained by joining the first to the last
mentioned point, creating AC.
C
C
B
A
a
c
b
b
c=a+b
A
B
(a) v1 = 6 m·s-1 East
v2 = 8 m·s-1 North
v3 = 6 + 8
2
8
2
θ
= 10 m ⋅ s -1 at 37o T
v = 40 + 30
2
30
2
3
= 50 m ⋅ s at 233 T
−1
Conceptual Physics – 3rd Edition – Paul Hewitt
tanθ = 8/6 =
θ = 53°
o
4
3
6
40
(b) v1 = 40 m·s-1 at 270° T
v2 = 30 m·s-1 at 180° T
B
a
θ
3
tanθ = 30/40 =
4
θ = 36.9°
Chapter 3 – Projectile Motion
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Components of Vectors
We have already learned that two vectors acting at a
point can be replaced by a single vector called the resultant.
Also a single vector can be regarded as the resultant of
two vectors at right angles. These two vectors are known as
components.
The process of determining these components is known
as resolution.
e.g.: using the vector R;
R
θ
Now, construct the components:
R
Vertical
Component
θ
Horizontal Component
AC = AB + BC
C (Note bold letters indicate vectors)
N.B. AC=AB + BC or AC2=AB2 + BC2
A
θ
B Vertical component = BC = AC sinθ
Horizontal component = AB = AC cosθ
Conceptual Physics – 3rd Edition – Paul Hewitt
Chapter 3 – Projectile Motion
Page 4 of 9
Vector Problems
Find the resultant vectors in both magnitude and
direction for the following:
Resultant velocity if:
a)
Ball 10.5 m·s-1
Wind 2.5 m·s-1
b)
wind
3 m·s-1
c)
Boat 10 m·s-1
3 m·s-1
5 m·s-1
d)
Conceptual Physics – 3rd Edition – Paul Hewitt
Plane Wind 50 m·s-1
300 m·s-1
Chapter 3 – Projectile Motion
Page 5 of 9
Vector Solutions
a) 10.5 m·s-1 North and 2.5 m·s-1 North;
2.5 m·s-1
13 m·s-1 North
10.5 m·s-1
b) 3 m·s-1 at 90° T and 10 m·s-1 at 0° T;
a = 3 m·s-1 at 90 °T
b = 10 m·s-1 at 0° T
c
b
c= a+b
c = 10.4 m·s-1 at 17 ° T
73°
a
Conceptual Physics – 3rd Edition – Paul Hewitt
Chapter 3 – Projectile Motion
Page 6 of 9
c) 5 m·s-1 East and 3 m·s-1 North;
C
AC = 5 + 3 = 5.83 m ⋅ s
2
A
3 m·s-1
θ
5 m·s-1
2
 3
θ = tan   = 30.96
5
−1
B
−1
o
AC = 5.8 m·s-1 at 59° T
d) 300 m·s-1 at 90° T and 50 m·s-1 at 0° T;
C
A
50 m·s-1
θ
300 m·s-1
B
AC = 50 + 300 = 304 m ⋅ s
2
2
−1
 50 
o
θ = tan −1 
 = 9.46
 300 
AC = 304 m·s-1 at 80.5° T.
Conceptual Physics – 3rd Edition – Paul Hewitt
Chapter 3 – Projectile Motion
Page 7 of 9
Projectile Motion
Assume in the following calculations that there is no friction
between the object and the air.
A projectile is any object that is projected by some
means near the earth’s surface and unless it is projected
vertically, it follows a curved path.
Vertical Component of Displacement
Horizontal Component of Displacement
• The horizontal component of velocity is constant.
• Vertical component of velocity is changing (the
object is accelerating).
• The components are independent of each other.
• Their combined effect produces a curved path, a
trajectory. This path is parabolic.
Conceptual Physics – 3rd Edition – Paul Hewitt
Chapter 3 – Projectile Motion
Page 8 of 9
Satellite Motion
We know that the horizontal and vertical
components of projectile motion are independent.
This means that no matter how fast we throw a ball
from the same height (for the same angle), it will
always hit the ground in the same amount of time.
However, the faster we throw it the larger the
horizontal distance it will cover in that same time.
10 m.s-1
5 m.s-1
If we take the curvature of the
Earth into account, we can see
that an object could be
constantly falling toward Earth,
without ever hitting it. The
object would have to be thrown so
fast that it falls toward Earth with a
trajectory matching the Earth’s curvature.
Conceptual Physics – 3rd Edition – Paul Hewitt
Chapter 3 – Projectile Motion
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