Solution 2.66
y
x
A8°
30 m
ff
Solution 2.67
Set up x–y coordinates with origin at A.
0
25cos
x
v
θ
=
y
Solution 2.68
x–y coordinates with origin at release point:
0
0
sin 12 sin
==
x
vv
θθ
y
Solution 2.69
y
A
D
x
θ
υ
0
80′
100′
0
40′
y
Solution 2.70
200 m/s υ
x0
= 200 cos 60° = 100 m/s
Solution 2.71
Set up x–y axes at A, target at B:
()
0
.: cos30
−= °
B
xeqx v t
Solution 2.72
22
1g
gsin g
R
θ
=
Solution 2.73
y
x
y
2g
60°
Solution 2.74
=°=
≤=°−=
2g cos 60 g
2 sec 2g sin 60 g 0.732g
ft
x
y
a
ta
()
2
2
2
2 sec g
2 64.4 ft/sec
1
2
y
xx
x
y
ta
vv
>=−
==
x, ft
y
, ft
583 ft = ymax
600
400
Transition
Solution 2.75
()
0
00
22
@: cos cos
11
xf
xx vt BR v t
αθ
=+ =
1
sin
2cos sin tan 1 0
1
t
an2 tan
1
2tan ta
θ
θθ α θ
θα
θ
−
−+=
=−
=−
()
11
180 tan
ntan
180 90 90
90
2
αα
αα
α
θ
−
=−
=°−°−=+
°+
∴
=
Specific results:
0, 45
30 , 60
45 , 67.5
αθ
αθ
αθ
==°
=° =°
=° = °
B
Solution 2.76
g
xx
akvj akv
=− − ∴=−
=− −
0
0
0
00
0cos
kt
xx
xt
kt
x
xkt kt
dx
vve
dt
dx v e dt
vv
θ
−
−
−−
==
=
0
0
sin
1g g
yy
kt
vv e v e
kk kk
kk
θ
−
=+ −= + −
x
y
k
Solution 2.77
22
13.5 2
45 5.30 ft/sec
13.5 2
vr
π
θ
π
== =
Solution 2.78
2
22 2 2 2
22.2 6.17 m/s
80
== =
n
v
a
ρ
Solution 2.79
a1: Speed is increasing, no path curvature.
Solution 2.80
2
=
n
v
a
ρ
Solution 2.81
2
215 km/hr or 59.7 m/s
=
v
c
Solution 2.82
2
2
,
== =
nn
v
aa v a
ρ
Solution 2.83
()
25 15 0.1745 rad
0.22
t
π
θ
Δ= − =
Δ
Δυ
t
Solution 2.84
max 40 km/hrv= at S = 60 m
The maximum acceleration for the sprinter will occur the instant he reaches his top speed
at the 60-m mark. At this location, he will experience his maximum tangential component
Solution 2.85
0
50 100
:12
=+ = +
tt
vv at a
Solution 2.86
2m/s
===
AB
vvv
Solution 2.87
50 100
4.75 , 163.0 m
0.6
nB
B
a
ρ
ρ
== =
−
Δυ
Solution 2.88
−−
vv
Halfway through time interval, 4.5 m/s=
v
1
2
2
2
22 2
2
2
4.5
1.5 0.060
1.5 m/s
=+= +
==
ptn
pt
aaa
a
a
Solution 2.89
()
cos
0.0260 m/s
==
=
vr R
θ
γ
θ
Solution 2.90
22
2
2
=+
nt
aa
a
υ
at = 1.5 m/s2
t
υ
0
θ
g
t
0
Solution 2.91
a
a
2
20 3.6 5.56 m/s
a
==
t
a
Solution 2.92
0165mphv= OR 242 ft/sec
θ
=12
At Launch:
g sin 32.2 sin12
t
a
θ
=− =− °
0
0
cos 242 cos12 237 ft/sec
0
x
t
vv
a
θ
== =
=
g
n
ϕ
Solution 2.93
φ
ρ
°−
== = −
00
22
2
242sin12 32.2
t
an 0.213 0.1360
237
yy
y
x
n
vtt
v
0.452sec 2.67 sec
tt
==
Solution 2.94
The radius of Jupiter is
37
139 822
2
2.41m/s
=
2
27
v
Solution 2.95
ρ
′
=
Bothtravel at320
constant speed
B
76.1
320 33.79sec
88.5
AA
AA
BB
BB
vv
vv
π
Car B Crosses DD First:
() ( )
76.1 41.3 3.79 25.9 ft
AA B
vt t
δδ
=−= −→=
Solution 2.96
22 2 2
y
υ
a
n
Solution 2.97
The time tup to apex is found from
vv t
=− = °−
22
68.1 149.7 ft
t
v
ρα
== =
Solution 2.98
2
22 2
1
19.62 95.3 97.3m/s
95.3
tan 78.4 168.4
19.62
−
=+=
===−
x
a
θθ
s
Solution 2.99
250 mm
2
29.7 mm/s
=
t
a
a
Solution 2.100
Accel. directed from A to N is
Solution 2.101
21
1
10 , in.
ykxk
−
−= =−
y
x = 5″
υ = 17970 km/hr