### Deriving the Range Equation

```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/s2
– Caused by the unbalanced force of gravity acting
on the object
ymax
x
x
R = 2x
ymax
vi
q
x
x
R = 2x
vi
viy = vi sin q
q
vx = vi cos q
ymax
vi
q
viy =
vi sin q
x
vx =vi cos q
x
R = 2x
In the x-direction:
=  , where t = time to top of path
= ( cos )t

=
cos
ymax
vi
q
x
x
R = 2x
In the y-direction:
=  + , ℎ  = 9.81 / 2
ymax
vi
q
x
x
R = 2x
In the y-direction:
=  + , ℎ  = 9.81 / 2
At the top of the path, vfy = 0
ymax
vi
q
x
x
R = 2x
So in the y-direction:
0 =  +
Substituting in

=
cos
ymax
vi
q
x
x
R = 2x
So in the y-direction:
0 =  +
Substituting in

=
cos
Now, we have

0 =  +
cos
Remember that the initial
velocity in the y-direction
= vi sin q
vi
viy =
vi sin q
q
vx = vi cos q
ymax
vi
q
x
x
R = 2x
So we go from

0 =  +
cos
To

0 =  sin  +
cos
The whole point here is to solve for x . . .
• − sin  =

cos
• (− sin )  cos  =
2sin  cos )
• (−
•
−
2
sin  cos

=x

(
cos
=
cos )
Remember that the range, R, = 2x
−2 2 sin  cos
= 2x = R

Double-angle theorem from trig:
2 sin  cos  = sin 2
so
−2 sin 2

=R
(Q.E.D.)
```