Maximum Range of a Projectile Launched from a Height—C.E. Mungan, Spring 2003
reference: TPT 41:132 (March 2003)
Find the launch angle q and maximum range R of a projectile launched from height h at speed u.
y
u
T
q
h
R
x
The basic equations of kinematics at the landing point after flight time T are
0 = h + u y T - 12 gT 2
(1)
R = u xT
(2)
vertically and
horizontally. Substitute Eq. (2) for T into (1) and convert from rectangular to polar components
to get
gR 2
- R sin 2q .
u2
(3)
1
R
fi q = tan -1 .
2
h
(4)
h (1 + cos 2q ) =
Maximize R by differentiating this expression with respect to q and putting dR / dq = 0 to obtain
an expression for the optimum launch angle,
tan 2q =
R
h
This implies cos 2q = h ( h 2 + R 2 ) -1/ 2 and sin 2q = R ( h 2 + R 2 ) -1/ 2 . Substitute these into Eq. (3)
to obtain the maximum range,
h=
gR 2 u 2
2u 2 2 g
fi R =
2u 2 Ê
u2 ˆ
Áh + ˜ .
2g ¯
g Ë
Equations (4) and (5) can be normalized for plotting purposes in terms of
(5)
R0 ∫
u2
,
g
(6)
the maximum range for a surface-to-surface projectile (i.e., when h = 0 ), to get the normalized
range
R
h
= 1+ 2
R0
R0
(7)
at a launch angle of
1
Ê R ˆ
q = sec -1 1 + 0 .
Ë
h¯
2
(8)
These two equations are plotted below as a function of the normalized launch height h / R0 .
5
45
40
normalized
range
4
35
30
3
25
20
2
launch angle
(degrees)
15
1
10
0
2
4
6
8
10
12
normalized launch height
As expected, R = R0 and q = 45˚ when h = 0 . In the other limit, R µ h 1/ 2 and q Æ 0˚ as h Æ • .
More reasonably, notice that if you launch from a height equal to 1.5 times your surface range,
you can get the projectile to go twice as far, provided you launch it at 26.6˚ (half of a 3–4–5
triangle angle).