# Deriving the Range Equation

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14 Νοε 2013 (πριν από 3 χρόνια και 1 μήνα)

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Deriving the Range Equation

Or, how to get
there

from
here

Keep in mind . . .

Horizontal velocity REMAINS CONSTANT

No net force is acting horizontally so there is no
horizontal acceleration

Vertical velocity CHANGES

Acceleration due to gravity, ~9.81 m/s
2

Caused by the unbalanced force of gravity acting
on the object

y
max

x

x

R = 2x

y
max

x

x

R = 2x

q

v
i

q

v
i

v
iy

= v
i
sin
q

v
x

= v
i

cos

q

y
max

x

x

R = 2x

q

v
i

v
iy

=

v
i
sin
q

v
x

=v
i

cos

q

In the x
-
direction:


=


, where t = time to top of path


=

(

𝑖

cos
𝜃
)
t

𝑜


=



𝑖
cos
𝜃

y
max

x

x

R = 2x

q

v
i

In the y
-
direction:


=

𝑖
+
𝑔
,
ℎ𝑒 𝑒

𝑔
=
9
.
81

𝑚
/

2

y
max

x

x

R = 2x

q

v
i

In the y
-
direction:


=

𝑖
+
𝑔
,
ℎ𝑒 𝑒

𝑔
=
9
.
81

𝑚
/

2

At the top of the path,
v
fy

= 0

y
max

x

x

R = 2x

q

v
i

So in the y
-
direction:

0
=

𝑖
+
𝑔

Substituting in


=



𝑖
cos
𝜃

y
max

x

x

R = 2x

q

v
i

So in the y
-
direction:

0
=

𝑖
+
𝑔

Substituting in


=



𝑖
cos
𝜃

Now, we have

0
=

𝑖
+
𝑔


𝑖
cos
𝜃

q

v
i

v
iy

=

v
i
sin
q

v
x

= v
i

cos

q

Remember that the initial
velocity in the y
-
direction

=

v
i

sin
q

y
max

x

x

R = 2x

q

v
i

So we go from

0
=

𝑖
+
𝑔


𝑖
cos
𝜃

To

0
=

𝑖
sin
𝜃
+
𝑔


𝑖
cos
𝜃

The whole point here is to solve for x . . .

𝑖
sin
𝜃
=


𝑣
𝑖
cos
𝜃

(

𝑖
sin
𝜃
)

𝑖
cos
𝜃
=


𝑣
𝑖
cos
𝜃
(

𝑖
cos
𝜃
)

(

𝑖
2
sin
𝜃
cos
𝜃
)
=
𝑔

𝑣
𝑖
2
sin
𝜃
cos
𝜃


= x

Remember that the range, R, = 2x

𝑣
𝑖
2
2

sin
𝜃
cos
𝜃


=
2
x =
R

Double
-
angle theorem from trig:

2
sin
𝜃
cos
𝜃
=
sin
2𝜃

so

𝑣
𝑖
2
sin
2
𝜃


=
R

(Q.E.D.)