s
g
Solution 2.38
2
g
tv
dv
akv
dt
=− =
1tan g
v
k
90 2.54
990 14.06
1000 14.19
100
Solution 2.39
2
(
a) 2 m/s constant==a
W
ith 250 3.6 69.4 m/s, we have
==
v
Solution 2.41
()
2
2
0
2
or 2
x
vdv adx d v adx
==
Solution 2.42
First, determine B’s acceleration time:
16.99s
=
t
Distance traveled by A in that time:
()
()
==
3
100 10 16.99 472 m
3600
A
d
A
C
B
Solution 2.43
m = 5 kg
µ = 0.40
k = 150 N/m
x0 = 200 mm
x
kµ
x0
min
0.1308 m or 130.8 mm
x
=
(mass will move left)
gg
xv
kvdv k
a
xxdxvdv
µµ
=− = → − =
222
1g
k
f
Solution 2.44
µ
=== =
0
5 kg, 0.40, 150 N/m, 500 mmmkx
From 2.43, =
min 130.8mmx to cause motion
For an initial stretch (motion to the left), gk
ax
m
µ
=−
2mg 2(5)(9.81)(0.40) 150(0.5)
kx
µ
−−
µ
150
ff
Solution 2.45
C
h
Solution 2.46
a
u
= –g – kυ
2
a
d
= –g + kυ
2
() ()
2
2
0
2 g 2 0.0035 32.2 0.0035 255
++
=84.3 ft/secv (use Article C.10)
v
tv
()
max
gg 32.2
32.2 0.0035
max
0
2
1 0.0035 32.2
y
v
−
Solution 2.47
()
5.1 0.4 5
0.02
+−
Δ
==
Δ
av
iji
r
vt
Solution 2.48
Solution 2.49
00
00
xy
() ( )
0
2
2
0.5 0.35 in./sec
0 0.5 0.35 0.5 0.175 in./sec
x
xx x
at
vv atdt tdt t t
=−
=+ =+ − = −
3.0
2.0
x–Coordinate, in.
1.0
0.0234567
y
y
Solution 2.50
22
111
()
()
2
=+= +−=
axy
Solution 2.51
2
600 mi/hr 880 ft/sec
30.8 ft/sec
;
y
v
a
vv at
=≡
=−
=+
Solution 2.52
()
00
max
@: 0 cos 1
sin 2.45
gg
xx vt B R v t
vv
R
θ
=+ =+
°
==
Solution 2.53
Angle
θ
for maximum range and, hence, minimum u is 45
From sample Prob. 2.6 range is
22
3
uu
Solution 2.54
θ
== °=
0
0
cos 45 cos 40 34.5 ft/sec
x
vv
y
Solution 2.55
60
θ
=°
AC
3.52 sec
AD
t∴=
Solution 2.56
()
0
22
00
0
19.81
g : 1.06 sin
22
y
yy
yy vt t v t t
θ
=+ − = −
y
x
AB
υ
0
θθ
υ
0
160′
d
Solution 2.58
y
x
s
B
θ
9 m
1.75 m
50°A4 m
12 5
12.5 m/s
• Impact:
111 1
54.915
−−− −
y
Solution 2.59
x
A
C
θ
1.65 m
11 m
DART
014 m/s=v, there is a 215 ms delay before the throw.
• Dart:
• Apple:
1.65 11 tan 9 2.07 m
++=→=
dd
θ
Solution 2.60
For y-motion, 2
g9.81 m/s==
y
a
x
u = 200 km/hr
x
θ
30×3 = 90′
Solution 2.61
2
222
22
g22
xuuy
Solution 2.62
y
ux
A
• Miss Corner B:
• Miss Corner C:
Solution 2.63
A
11.7 m
0.9 + 0.15 = 1.05 m
2.55 m
B
CS
x
υ
A
()
0
T
a
0 : , 11.7
2.55 9.81 , 0.721s
2
A
:
t
== =
==
xx B
cc
a
xvt vt
tCt
Solution 2.64
A
B
CD
y
υ
60‘
θ
0
hr
• Check A:
• Check B:
First impact point is 42.2 ft up in the tree at B.
Solution 2.65
()
42 m/s, and 8 :
==°
v
a
θ