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CHAPTER 7
PROJECTILES
7.1
No Air Resistance
We suppose that a particle is projected from a point O at the origin of a coordinate system,
the
y
axis being vertical and the
x
axis directed along the ground.
The particle is projected in
the
xy
plane, with initial speed
V
0
at an angle
α
to the horizon.
At any subsequent time in its
motion its speed is
V
and the angle that its motion makes with the horizontal is
ψ
.
The initial horizontal component if the velocity is
V
0
cos
α
, and, in the absence of air
resistance, this horizontal component remains constant throughout the motion.
I shall also
refer to this constant horizontal component of the velocity as
u
.
I.e.
u
=
V
0
cos
α
=
constant
throughout the motion.
The initial vertical component of the velocity is
V
0
sin
α
, but the vertical component of the
motion is decelerated at a constant rate
g
.
At a later time during the motion, the vertical
component of the velocity is
V
sin
ψ
, which I shall also refer to as
v
.
In the following, I write in the left hand column the horizontal component of the equation of
motion and the first and second time integrals; in the right hand column I do the same for the
vertical component.
Horizontal.
Vertical
0
=
x
g
y

=
7.1.1
a
,
b
α
=
=
cos
0
V
u
x
gt
V
y

α
=
=
sin
0
v
7.1.2
a
,
b
x
V t
=
0
cos
α
y
V t
gt
=

0
1
2
2
sin
α
7.1.3
a
,
b
The two equations 7.1.3
a
,
b
are the parametric equations to the trajectory. In vector form,
these two equations could be written as a single vector equation:
r
V
g
0
=
+
t
t
1
2
2
.
7.1.4
Note the + sign on the right hand side of equation 7.1.4.
The vector
g
is directed downwards.
The
xy
equation to the trajectory is found by eliminating
t
between equations 7.1.3
a
and
7.1.3
b
to yield:
y
x
gx
V
=

tan
cos
.
α
α
2
0
2
2
2
7.1.5
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Now, rewrite this in the form
( 29 ( 29
.
2
y
x
x

=

Add to each side (half the coefficient of
x
)
2
in order to "complete the square" on the left hand
side, and, after some algebra, it will be found that the equation to the trajectory can be written
as:
( 29 ( 29
,
4
2
B
y
a
A
x


=

7.1.6
where
A
V
g
V
g
=
=
0
2
0
2
2
2
sin
cos
sin
,
α
α
α
7.1.7
B
V
g
=
0
2
2
2
sin
,
α
7.1.8
and
a
V
g
=
0
2
2
2
cos
.
α
7.1.9
Having rearranged equation 7.1.5 in the form 7.1.6, we see that the trajectory is a
parabola
whose vertex is at (
A
,
B
).
The
range on the horizontal plane
is 2
A,
or
V
g
0
2
2
sin
.
α
The
greatest range on the horizontal plane is obtained when sin 2
α
= 1, or
α
= 45
o
.
The greatest
range on the horizontal plane is therefore
V
g
0
2
/ .
The
maximum height
reached is
B
, or
V
g
0
2
2
2
sin
.
α
The
distance between vertex and focus
is
a
, or
V
g
0
2
2
2
cos
.
α
The focus is above
ground if this is less than the maximum height, and below ground if it is greater than the
maximum height.
That is, the focus is above ground if cos
sin
.
2
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