Solution 2.193
A
y
20°
45°
υ
A
, a
A
Solution 2.194
υ
1
45°
y
u =
200 ft/sec
y
Solution 2.195
/
(a) WB W B
vvv=−
()()
20 sin35 cos35 16 sin15 cos15
15.61 0.928 mi/hr
ij ij
ij
= − °− ° − °− °
=− −
20 cos 40 sin 40
ijvj
′′′
=− °− °−−
Solution 2.196
/
2
so
+=+
x
PABPB
vvvv
y
‘
y
B
y
υ
Px
Solution 2.197
a
B/A
= 2.4 m/s
2
() ()()
11 1
1.864 2.32 1.439 m/s
33 3
AB B A
vv at v
== = →= ↓
Solution 2.198
Carrier deck has a constant velocity, so may be used as an inertial coordinate base.
Velocity of aircraft relative to carrier is
AA
Solution 2.199
υ
r
υ = 30 m/s
2
60°
a
θ
t
θ
2
6.93 m/s
26.0 m/s
θ
=
=
[
]
θ
−
r
6.93
n
n
ρ
Solution 2.200
3.25st=
() () ()
() ()
()
23
23
2
2
10.25 1.75 0.45 10.25 3.25 1.75 3.25 0.45 3.25 36.3mm
10.25 3.5 1.35 10.25 3.5 3.25 1.35 3.25 7.37 mm/s
3.5 2.7 3.5 2.7 3.25 5.28mm/s
xttt
xtt
xt
=+−= + − =
()
14.65 4.96 14.65 4.96 3.25 1.47 mm/s
yt
=−=− =−
22 2 2
7.37
7.37 1.47 0.981 0.1957
7.51
−
== → = −
tt
x
ij
v
eeij
v
Thus, 0.1957 0.981
n
eij=− −
()()
2
5.28 4.96 0.981 0.1957 4.20 mm/s
=⋅ =− − ⋅ − → =−
t
t t
aae i j i j a
7.51 9.57 mm
ρρ
=→== →=
n
vv
aa
n
ÿ
Solution 2.201
23 2
10.25 1.75 0.45 mm & 6.32 14.65 2.48 mmxttt y tt=+− =+−
AT 3.25s :t=
() ()
=+ − =
23
2
2
10.25 3.5 3.25 1.35 3.25 7.37 mm/s
x
36.3
x
θθ
=+= °+ °→= +
cos sin cos37.3 sin37.3 0.795 0.607
rr
eij i je ij
2
r
r r
rr r
r
()
2
2
0.743 0.607 0.795 0.451 0.591 mm/s
aae i j a i j
θθ θ
θ
=⋅ =− − ⋅− + → =−
==− − + →= −
Solution 2.202
Horizontal motion:
()
0
0
: 0 130 9.65 1255 ft
x
xx vtx
=+ =− =−
2
gg R
=
()
22
00
Substitute numerical values and obtain
or =6048km
R
Solution 2.204
425 mm
A
S
=
Differentiate WRT time:
1
12
11
−
−
B
C
B
S
B
Solution 2.205
From the solution to Prob. 2.204, the velocity constraint equation is
1
12
22 2
22
AA A AA B B A B
−
−
The next derivative is
() ()
13
1
2
22
3
250 2 250
−−
−
−
+− + + − ×
AA AA BB B B
AB AA B B
SS Sv S v
13
and 46.8mm/s P
r
ob
B
v
=
−
()
. 2 / 204 , along wit
h
AB
Solution 2.206
2
cos , sin , cos
ttpt
θ θ ωθ θω ωθ θω
==− =−
()
885 m, 222 m, 912 m, 14.10
98.3 m/s, 19.46 m/s down
xy
xyr
vv
θ
====°
==−
–10
–9.51
024689101214
a
θ
Solution 2.208
gcos
gcos
d
dl
θθ
θ
θθ
θ
θ
==
θ
l
Then
1/2
2
2g sin
d
θ
θ
θθ
==+
θ
2
0.4
Solution 2.209
−
−+
=−=−
2
2
2( )
2
1g g
ln 1
kyh
dv
k
kv yh v e
=99.8 ft/sec
t
v
Without drag:
()()
′= = =2g 2 32.2 200 113.5 ft/secvh
200
100
00 10050
y
, ft
υ, ft/sec
g
y
Solution 2.210
4.50 4.50
Tv
−
Where 250 kN,T= s = distance in metres, 16000m= tons (metric) (1 metric ton=1000 kg)
speed inm/sv=
or if v = speed in knots & s = distance in nautical miles, then
()
9.00 1852s/16000 1.042s
1mi
max
250 3.6
1 14.49 1
4.50 1.852
11.66 knots
14.49 knots
ve e
v
v
−× −
=− =−
=
=
8
6
Solution 2.211
gcos
t
dv k
a
v
θ
== −
dk
max
110.4 @ 0.802 sec
3.79 rad/sec @ 0.324 sec
t
t
θ
θ
=°=
==
Values used: 2
1.8 ft
slugs, g 32.2
32.2 sec
m==
1
v
0123
t, sec
–1
–0.5
0
1
3.5
4
45
Solution 2.212
Set up x–y coordinates @ A:
Let coordinates of B be (R, −h).
y
Solution 2.213
300 m
y
A′
Aθ
80
S
B
, m
Solution 2.214
0
65 m/s
=
v
With Drag:
x
vx=and y
vy=;x
ax= and y
ay=
00
(0) 0 and (0) cos (0) 0, (0) sin
xxv yyv
θθ
== ==
These two second-order differential equations must be numerically integrated since they
are nonlinear and coupled.
x
70
30
x–Coordinate, m
Particle trajectory