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Thus, one root is related to the other (generically labeled
h'
and
h
) by
h'
=
H
–
h
. Its
numerical value is
' 40cm 10 cm
30 cm.
h
=−
=
(c) We wish to maximize the function
f
=
x
2
= 4
h
(
H
–
h
). We differentiate with respect to
h
and set equal to zero to obtain
48
0
2
df
H
Hh
h
dh
=
¡
=
or
h
= (40 cm)/2 = 20 cm, as the depth from which an emerging stream of water will
travel the maximum horizontal distance.
65. (a) Since Sample Problem 148 deals with a similar situation, we use the final
equation (labeled “Answer”) from it:
0
2
for the projectile motion.
vg
hv
v
=
¡
=
The stream of water emerges horizontally (
θ
0
= 0° in the notation of Chapter 4), and
setting
y
–
y
0
= –(
H
–
h
) in Eq. 422, we obtain the “timeofflight”
2(
)
2
()
.
tH
h
gg
−−
==
−
−
Using this in Eq. 421, where
x
0
= 0 by choice of coordinate origin, we find
0
2(
)
2
2
(
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Projectile Motion

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