101
*12–96.
It is observed that the skier leaves the ramp Aat an angle
with the horizontal. If he strikes the ground at B,
determine his initial speed and the time of flight .tAB
vA
uA=25°
SOLUTION
A
:
+
B
s=v0t
4 m
vA
100 m
A
3
4
5
uA
Ans:
102
12–97.
4 m
vA
100 m
B
A
3
4
5
uA
It is observed that the skier leaves the ramp Aat an angle
with the horizontal. If he strikes the ground at B,
determine his initial speed and the speed at which he
strikes the ground.
vA
uA=25°
SOLUTION
Coordinate System: coordinate system will be set with its origin to coincide
with point Aas shown in Fig.a.
Substitute Eq. (1) into (2) yieldS
yB =yA+(vA)y t+1
A
+c
B
x–y
103
12–97. Continued
Using this result,
And
Ans:
104
12–98.
SOLUTION
Vertical Motion:The vertical component of initial velocity is .For the ball
to travel from Ato B, the initial and final vertical positions are and
, respectively.
Horizontal Motion:The horizontal component of velocity is .For the ball
to travel from Ato B, the initial and final horizontal positions are and
(s0)x=0
(v0)x=vA
A
+c
B
sy=(s0)y+(v0)yt+1
2(ac)yt2
sy=3ft
(s0)y=7.5 ft
(v0)y=0
Determine the horizontal velocity of a tennis ball at Aso
that it just clears the net at B. Also, find the distance swhere
the ball strikes the ground.
vA
21 ft
s
7.5 ft
3ft
vA
C
A
B
Ans:
SOLUTION
dv=a dt
Ls
0
ds =
Lt
0
2.5t2 dt
12–99.
The missile at A takes off from rest and rises vertically to B,
where its fuel runs out in 8 s. If the acceleration varies with
time as shown, determine the missile’s height hB and
speed vB. If by internal controls the missile is then suddenly
pointed 45° as shown, and allowed to travel in free flight,
determine the maximum height attained, hC, and the
range R to where it crashes at D.
R
A
B
D
C
v
B
h
B
45
h
C
Ans:
106
*12–100.
The projectile is launched with a velocity . Determine the
range R, the maximum height hattained, and the time of
flight. Express the results in terms of the angle and .The
acceleration due to gravity is g.
v0
u
v
0
SOLUTION
s=s0+v0 t+1
2 ac t2
A
+c
B
s=s0+v0t
A
+
:
B
y
x
R
h
v0
u
Ans:
SOLUTION
Horizontal Motion:
Vertical Motion:
Substituting Eq.(1) into (2) yields:
12–101.
The drinking fountain is designed such that the nozzle is
located from the edge of the basin as shown. Determine the
maximum and minimum speed at which water can be
ejected from the nozzle so that it does not splash over the
sides of the basin at B and C.
A
BC
v
A
50 mm
100 mm
250 mm
40
1
00 mm
108
SOLUTION
A
Also,
Vertical Motion. Here,
(
vA
)
y
=10 sin
uA
c
,
(s
A
)
y
=(s
B
)
y
=0
and ay=9.81 m
>
s
2
T
Also
Substitute Eq. (1) into (3)
Using the trigonometry identity
sin 2
u
A=2 sin
u
A cos
u
A,
this equation becomes
sin 2
2
2
12–102.
If the dart is thrown with a speed of 10 m
>
s, determine the
shortest possible time before it strikes the target. Also, what
is the corresponding angle
u
A at which it should be thrown,
and what is the velocity of the dart when it strikes the
target?
B
4 m
AvA
u
A
109
Since the shorter time is required, Eq. (1) indicates that smaller u
A
must be choosen.
Thus
A=11.55°=11.6°
and
Substitute the result of
uA
into Eq. (2) and (4)
Thus, the magnitude of
vB
is
12–102. Continued
Ans:
110
12–103.
If the dart is thrown with a speed of 10 m
>
s, determine the
longest possible time when it strikes the target. Also, what is
the corresponding angle
u
A at which it should be thrown,
and what is the velocity of the dart when it strikes the
target?
B
4 m
AvA
u
A
SOLUTION
Coordinate System. The origin of the
x@y
coordinate system will be set to coincide
with point A as shown in Fig. a.
Horizontal Motion. Here,
(vA)x=10 cos
u
A S,
(sA)x=0
and
(sB)x=4 m S.
Also,
(vB)x=(vA)x=10 cos
Vertical Motion. Here,
(
vA
)
=10 sin
uA
c
,
(s
)
=(s
)
=0
and
Since
t = 0,
then
Also,
(
)
Substitute Eq. (1) into (3)
Using the trigonometry identity
sin 2
u
A=2 sin
u
A cos
u
A,
this equation becomes
sin 2
2
2
111
Since the longer time is required, Eq. (1) indicates that larger u
A
must be choosen.
Thus,
and
Substitute the result of
uA
into Eq. (2) and (4)
12–103. Continued
Ans:
112
*12–104.
The man at Awishes to throw two darts at the target at Bso
that they arrive at the same time. If each dart is thrown with
a speed of determine the angles and at which
they should be thrown and the time between each throw.
Note that the first dart must be thrown at then
the second dart is thrown at uD.
uC17uD2,
uD
uC
10 m>s,
SOLUTION
From Eq. (1),
The two roots are Ans.
uD=14.7°
5=20.39 sin ucos u
(:
+)s=s0+v0t
B
5m
AD
C
u
C
u
D
Ans:
113
12–105.
The velocity of the water jet discharging from the orifice can
be obtained from ,where is the depth of
the orifice from the free water surface.Determine the time
for a particle of water leaving the orifice to reach point B
and the horizontal distance xwhere it hits the surface.
h=2mv=22gh
SOLUTION
Coordinate System: The x–y coordinate system will be set so that its origin coincides
x-Motion: Here,,,,and .Thus,
t=tA
xB=xxA=0(vA)x=vA=6.264 m>s
1.5 m
2m A
x
B
vA
114
12–106.
The snowmobile is traveling at 10 m
>
s when it leaves the
embankment at A. Determine the time of fl ight from A to B
and the range R of the trajectory.
SOLUTION
(
S
+
)
sB=sA+vA
t
40
3
4
5
R
B
A
115
12–107.
SOLUTION
A
:
+
B
s=s0+v0t
The fireman wishes to direct the flow of water from his hose
to the fire at B. Determine two possible angles and at
which
this can be done.Water flows from the hose at
.vA=80 ft>s
u2
u1
20 ft
A
u
B
vA
116
*12–108.
The baseball player Ahits the baseball at and
from the horizontal. When the ball is directly
overhead of player Bhe begins to run under it. Determine
the constant speed at which Bmust run and the distance d
in order to make the catch at the same elevation at which
the ball was hit.
uA=60°
vA=40 ft
>
s
SOLUTION
Vertical Motion:The vertical component of initial velocity for the football is
Horizontal Motion:The horizontal component of velocity for the baseball is
vA=40 ft/s
A
vA
A
θ
BC
15 ft d
Ans:
117
12–109.
The catapult is used to launch a ball such that it strikes the
wall of the building at the maximum height of its trajectory.
If it takes 1.5 s to travel from Ato B, determine the velocity
at which it was launched, the angle of release and the
height h.
u,vA
3.5 ft
h
A
B
v
A
18 ft
u
SOLUTION
To
solve, first divide Eq. (2) by Eq. (1), to get .Then
u
(
:
+)s=n0t
Ans:
12–110.
An automobile is traveling on a curve having a radius of
800 ft. If the acceleration of the automobile is ,
determine the constant speed at which the automobile is
traveling.
5ft>s2
SOLUTION
Acceleration: Since the automobile is traveling at a constant speed, .
at=0
Ans:
119
12–111.
SOLUTION
Acceleration:Since the speed of the race car is constant, its tangential component of
acceleration is zero, i.e., .Thus,
at=0
Determ
i
ne t
h
e max
i
mum constant spee
d
a race car can
have if the acceleration of the car cannot exceed
while rounding a track having a radius of curvature of
200 m.
7.5 m>s2
Ans:
120
SOLUTION
1.5s ds =
v dv
*12–112.
A boat has an initial speed of 16 ft
>
s. If it then increases its
speed along a circular path of radius
r
= 80 ft at the rate of
v
#
= (1.5s) ft
>
s, where s is in feet, determine the time needed
for the boat to travel s = 50 ft.
Ans: