Solution 3.276
22
0: mg
:
yy
nn
FN
vv
Fm Nm
ρρ
=
=
=
=
Solution 3.277
2
Drop of A state state :
1
:1.8 1 cos60 2
()
TU TOmg mv
→
+= +
−°=
() ()( )
2
S
olution: 2.42m/s, 5.36 m/s
15.36 9.81 2.4 1 cos30 sin30 0
BB
vv
mm s
==

−−°+°=

0
Ny
y t
mg
nn
F
(a)
N
y
Solution 3.278
Results of Problem 3.241:
2
g1
r
Nominally,
()( )( )
3959 5280 1 7997
An
v

Δ= × −=
Actually,
()( )( )
()()
()()
3959 5280 1 755
3959 170 5280 3959 170 3959 700 sec
Aa
v


Δ= × −=

++++

Solution 3.280
(Roman numeral: process; Arabic number: state)
22
1
6
6
32.2
( ) 2.69ft/sec
v
→=
Solution 3.281
Solve cubic : 268ft/secor182.9 mi hr
v
′=
Solution 3.282
For a minimum escape orbit (parabolic) e = 1 and →∞a so from Eq. 3.47
T
hus, 35140 26140 9000 km/h
v
Δ= − =
υ
F
R
= 200 lb
F
1
–
2
ρυ
2
F
D
= C
D
S
)(
Solution 3.283
()
or 3.73 35.4 1.172 2 47.3ft
x
=+ +=
4‘
B
8‘
2‘
x
100 ft/sec
e2 = 0.3
1
2
34
Solution 3.284
Solution 3.285
0.5
0.289 m
o
Bo
ABo
S
SS
==
Generally:
222
2
L
O
A
90°-θ
S
B
S
A
120°
Am
B
m
30°
0
0
1
2
3
4
0.02 0.04 0.06
S
B
– S
Bo
, m
0.08 0.1 0.12 0.14 0.16
│υ
A
│ m/s
00.02 0.04 0.06
0.08 0.1 0.12 0.14 0.16
0
1
2
3
4
υ
B
m/s
Solution 3.286
2
(
b) Repeat for in direction (b) to
sin cos
obtain (1 sin )
s
s
s
FN
r
µ
θµ θ θω
=
+=+
y
Solution 3.287
()
/
2
2
:mgcos sin
cFCF
nn
aaa
FmaN mr a
θθ θ
=+
=− =+
()
()()
2
22
0
00
2cosgsin , 2singcos,
2sin g cos g
rd a dr a
r
θθ θ
θθθθθθθ
=− =+

Substitute into (2) and get
max
2
M
ax occurs when 0 soa sin g cos g 0
g
so 53.1 when .
2
g
2
=+−=
=° =
a
θθθθ
θ
2.5
1.5
Angular velocity theta dot, rad/s
0.5
0.0
01020
Angle theta, degrees
30 40 50 60
F
C
N
Solution 3.288
Define states
() ()
00
01
2
1
sin 80 sin5 6.97 ft/sec
30
t 0.376sec
79.7
3 6.97 0.376 16.1 0.376 3.34 ft
79.7 ft/sec
y
vv
y
vv
θ
==°=
==
=+ − =
==
21
3223
43
(2)
(7)
yy
x
xx
vv
vv
=
=
1
A
y
5°
80 ft/sec
e = 0.5
Solution 3.289
2
0
0.535
,1.069s
0.5
===
tt
θω
Solution 3.290
AB BA
t-momentum for baseball:
υ
B
‘
tn
(LOI)
(β = θ + α)
α
O
N
θ
15.12516
With slug, slug,
32.2 32 2
BB
AB
mm
==
20 26.5 46.5 , and initial attitude 3 .)
βθα
ο
=+= °+ = ° =
′
0.4 0.45 0.5
e
0.55 0.6
400
e