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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

+ uyT

-

1 2

gT 2

(1)

vertically and

R = uxT

(2)

horizontally. Substitute Eq. (2) for T into (1) and convert from rectangular to polar components to get

gR2

h(1 + cos2q) = u2 - Rsin2q .

(3)

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

fi

q

=

1 2

tan-1

R h

.

(4)

This implies cos2q = h (h2 + R2)-1/2 and sin2q = R (h2 + R2)-1/2 . Substitute these into Eq. (3) to obtain the maximum range,

h

=

gR2 2u 2

-

u2 2g

fi

R=

2u g

2

? ? ?

h

+

u2 2g

^ ~ ?

.

(5)

Equations (4) and (5) can be normalized for plotting purposes in terms of

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

R0 = 1 + 2 R0

(7)

at a launch angle of

q

=

1 2

sec-1?1 ?

+

R0 ^. h?

(8)

These two equations are plotted below as a function of the normalized launch height h / R0.

45

5

40

normalized

range

4

35

30 3

25

20 2

launch angle

15

(degrees)

10

1

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 ? h1/2 and q ? 0? as h ? ?. 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).

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