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:
𝑥 = 𝑣𝑥 𝑡, where t = time to top of path
𝑥 = (𝑣𝑖 cos 𝜃)t
𝑥
𝑠𝑜 𝑡 =
𝑣𝑖 cos 𝜃
ymax
vi
q
x
x
R = 2x
In the y-direction:
𝑣𝑓𝑦 = 𝑣𝑖𝑦 + 𝑔𝑡, 𝑤ℎ𝑒𝑟𝑒 𝑔 = 9.81 𝑚/𝑠 2
ymax
vi
q
x
x
R = 2x
In the y-direction:
𝑣𝑓𝑦 = 𝑣𝑖𝑦 + 𝑔𝑡, 𝑤ℎ𝑒𝑟𝑒 𝑔 = 9.81 𝑚/𝑠 2
At the top of the path, vfy = 0
ymax
vi
q
x
x
R = 2x
So in the y-direction:
0 = 𝑣𝑖𝑦 + 𝑔𝑡
Substituting in
𝑥
𝑡=
𝑣𝑖 cos 𝜃
ymax
vi
q
x
x
R = 2x
So in the y-direction:
0 = 𝑣𝑖𝑦 + 𝑔𝑡
Substituting in
𝑥
𝑡=
𝑣𝑖 cos 𝜃
Now, we have
𝑥
0 = 𝑣𝑖𝑦 + 𝑔
𝑣𝑖 cos 𝜃
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
𝑥
0 = 𝑣𝑖𝑦 + 𝑔
𝑣𝑖 cos 𝜃
To
𝑥
0 = 𝑣𝑖 sin 𝜃 + 𝑔
𝑣𝑖 cos 𝜃
The whole point here is to solve for x . . .
• −𝑣𝑖 sin 𝜃 =
𝑔𝑥
𝑣𝑖 cos 𝜃
• (−𝑣𝑖 sin 𝜃) 𝑣𝑖 cos 𝜃 =
• (−𝑣𝑖
•
−𝑣𝑖
2sin 𝜃 cos 𝜃)
2
sin 𝜃 cos 𝜃
𝑔
=x
𝑔𝑥
(𝑣𝑖
𝑣𝑖 cos 𝜃
= 𝑔𝑥
cos 𝜃)
Remember that the range, R, = 2x
−𝑣𝑖2 2 sin 𝜃 cos 𝜃
= 2x = R
𝑔
Double-angle theorem from trig:
2 sin 𝜃 cos 𝜃 = sin 2𝜃
so
−𝑣𝑖2 sin 2𝜃
𝑔
=R
(Q.E.D.)