The horizontal range is the displacement in horizontal direction. There is no acceleration involved in this direction. Motion is an uniform motion. It follows that horizontal range is transversed with the horizontal component of the projection velocity for the time of flight (T). Now,
x
=
u
x
t
⇒
R
=
u
x
T
=
u
cos
θ
x
T
x
=
u
x
t
⇒
R
=
u
x
T
=
u
cos
θ
x
T
where “T” is the time of flight. Putting expression for the time of flight,
⇒
R
=
u
cos
θ
x
2
u
sin
θ
g
⇒
R
=
u
cos
θ
x
2
u
sin
θ
g
⇒
R
=
u
2
sin
2
θ
g
⇒
R
=
u
2
sin
2
θ
g
(3)
Horizontal range, like time of flight and maximum height, is greater for greater projection speed.
For projection above ground surface, the range of the angle of projection with respect to horizontal direction, θ, is 0° ≤ θ ≤ 90° and the corresponding range of 2θ is 0° ≤ 2θ ≤ 180°. The “sin 2θ”, as appearing in the numerator for the expression of the horizontal range, is an increasing function for 0° ≤ θ ≤ 45° and a decreasing function for 45° ≤ θ ≤ 90° . For θ = 45°, sin 2θ = sin 90° = 1 (maximum).
In the nutshell, the range,R, increases with increasing angle of projection for 0° ≤ θ < 45°; the range,R, is maximum when θ = 45°; the range,R, decreases with increasing angle of projection for 45° < θ ≤ 90°.
The maximum horizontal range for a given projection velocity is obtained for θ = 45° as :
⇒
R
=
u
2
g
⇒
R
=
u
2
g
There is an interesting aspect of “sin 2θ” function that its value repeats for component angle i.e (90° – θ). For, any value of θ,
sin
2
(
90
0

θ
)
=
sin
(
180
0

2
θ
)
=
sin
2
θ
sin
2
(
90
0

θ
)
=
sin
(
180
0

2
θ
)
=
sin
2
θ
It means that the range of the projectile with a given initial velocity is same for a pair of projection angles θ and 90° – θ. For example, if the range of the projectile with a given initial velocity is 30 m for an angle of projection, θ = 15°, then the range for angle of projection, θ = 90°  15° = 75° is also 30m.
Equation of projectile motion renders to few additional forms in terms of characteristic features of projectile motion. One such relation incorporates range of projectile (R) in the expression. The equation of projectile motion is :
y
=
x
tan
θ

g
x
2
2
u
2
cos
2
θ
y
=
x
tan
θ

g
x
2
2
u
2
cos
2
θ
The range of projectile is :
R
=
u
2
sin
2
θ
g
=
u
2
2
sin
θ
cos
θ
g
R
=
u
2
sin
2
θ
g
=
u
2
2
sin
θ
cos
θ
g
Solving for
u
2
u
2
,
⇒
u
2
=
R
g
2
sin
θ
cos
θ
⇒
u
2
=
R
g
2
sin
θ
cos
θ
Substituting in the equation of motion, we have :
⇒
y
=
x
tan
θ

2
sin
θ
cos
θ
g
x
2
2
R
g
cos
2
θ
⇒
y
=
x
tan
θ

2
sin
θ
cos
θ
g
x
2
2
R
g
cos
2
θ
⇒
y
=
x
tan
θ

tan
θ
x
2
R
⇒
y
=
x
tan
θ

tan
θ
x
2
R
⇒
y
=
x
tan
θ
1

x
R
⇒
y
=
x
tan
θ
1

x
R
It is a relatively simplified form of equation of projectile motion. Further, we note that a new variable “R” is introduced in place of “u”.
If points of projection and return are on same level and air resistance is neglected, which of the following quantities will enable determination of the range of the projectile (R) :
(a) horizontal component of projection velocity
(b) projection speed and angle of projection
(c) vertical component of projection velocity
(d) speed at the highest point
The horizontal range is determined using formulae,
R
=
u
2
sin
2
θ
g
R
=
u
2
sin
2
θ
g
Hence, horizontal range of the projectile can be determined when projection speed and angle of projection are given. The inputs required in this equation can not be made available with other given quantities.
Hence, option (b) is correct.
Note : Horizontal range (R) unlike time of flight (T) and maximum height (H), depends on both vertical and horizontal motion. This aspect is actually concealed in the term "sin2
θ
θ
". The formula of horizontal range (R) consists of both "u sin
θ
θ
" (for vertical motion) and "u cos
θ
θ
" (for horizontal motion) as shown here :
⇒
R
=
u
2
sin
2
θ
g
=
2
u
2
sin
θ
cos
θ
g
=
2
u
cos
θ
u
sin
θ
g
⇒
R
=
u
2
sin
2
θ
g
=
2
u
2
sin
θ
cos
θ
g
=
2
u
cos
θ
u
sin
θ
g
⇒
R
=
2
u
x
u
y
g
⇒
R
=
2
u
x
u
y
g
A projectile is thrown with a velocity
6
i
+
20
j
m
/
s
6
i
+
20
j
m
/
s
. Then, the range of the projectile (R) is :
a
12
m
b
20
m
c
24
m
d
26
m
a
12
m
b
20
m
c
24
m
d
26
m
We shall not use the standard formulae as it would be difficult to evaluate angle of projection from the given data. Now, the range of the projectile (R) is given by :
R
=
u
x
T
R
=
u
x
T
Here,
u
x
=
6
m
/
s
u
x
=
6
m
/
s
We need to know the total time of flight, T. For motion in vertical direction, the vertical displacement is zero. This consideration gives the time of flight as :
T
=
2
u
y
g
=
2
X
20
10
=
4
s
T
=
2
u
y
g
=
2
X
20
10
=
4
s
Hence, range of the flight is :
⇒
R
=
u
x
T
=
6
X
4
=
24
⇒
R
=
u
x
T
=
6
X
4
=
24
Hence, option (c) is correct.