53. (a) From Table 6-1 and Eq. 6-16, we have
vt=2Fg
CρA =⇒CρA =2
mg
v2
t
where vt= 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. 6-14:
D=1
2CρAv2=1
22mg
v2
tv2=mg v
vt2
which yields D= (690)(360/60)2≈2×104N.
(b) We assume the mass of the ejection seat is roughly equal to the mass of the pilot. Thus, Newton’s
second law (in the horizontal direction) applied to this system of mass 2mgives the magnitude of
acceleration:
|a|=D
2m=g
2v
vt2
=18g.