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40. (a) From Table 61 and Eq. 616, we have
v
F
CA
mg
v
t
g
t
=
¡
=
2
2
2
ρ
where
v
t
= 60 m/s. We estimate the pilot’s mass at about
m
= 70 kg. Now, we convert
v
=
1300(1000/3600)
≈
360 m/s and plug into Eq. 614:
DC
A
v
mg
v
vm
g
v
v
tt
==
F
H
G
I
K
J
=
F
H
G
I
K
J
1
2
1
2
2
2
2
2
2
which yields
D
= (70 kg)(9.8 m/s
2
)(360/60)
2
≈
2
×
10
4
N.
(b) We assume the mass of the ejection seat is roughly equal to the mass of the pilot.
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 Spring '08
 Any
 Physics, Mass

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