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•12–5. A particle is moving along a straight line with the

acceleration a = (12t – 3t1/2) ft>s2, where t is in seconds.
Determine the velocity and the position of the particle as a
function of time. When t = 0, v = 0 and s = 15 ft.
Velocity:
A:
+ B dv = a dt
v t
L0 L0
dv = A 12t - 3t1>2 B dt
v t
v!0 = A 6t2 - 2t3>2 B 2
0
v = A 6t2 - 2t3>2 B ft>s Ans.
Position: Using this result and the initial condition s = 15 ft at t = 0 s,
A:
+ B ds = v dt
s t
L15 ft L0
ds = A 6t2 - 2t3>2 B dt
s 4 5>2 2 t
s!15 ft = a 2t3 - t b
5 0
4 5>2
s = a 2t3 - t + 15 b ft Ans.
5
12–6. A ball is released from the bottom of an elevator

which is traveling upward with a velocity of 6 ft>s. If the ball
strikes the bottom of the elevator shaft in 3 s, determine the
height of the elevator from the bottom of the shaft at the
instant the ball is released. Also, find the velocity of the ball
when it strikes the bottom of the shaft.
Kinematics: When the ball is released, its velocity will be the same as the elevator at
the instant of release. Thus, v0 = 6 ft>s. Also, t = 3 s, s0 = 0, s = -h, and
ac = -32.2 ft>s2.
1
A+cB s = s0 + v0t + a t2
2 c
1
-h = 0 + 6(3) + (-32.2) A 32 B
2
h = 127 ft Ans.
A+cB v = v0 + act
v = 6 + (-32.2)(3)
= -90.6 ft>s = 90.6 ft>s T Ans.
3
91962_01_s12-p0001-0176 6/8/09 8:06 AM Page 10
12–18. A car starts from rest and moves with a constant

acceleration of 1.5 m>s2 until it achieves a velocity of 25 m>s.
It then travels with constant velocity for 60 seconds.
Determine the average speed and the total distance traveled.
Kinematics: For stage (1) of the motion, v0 = 0, s0 = 0, v = 25 m>s, and

ac = 1.5 m>s2.
A:
+ B v = v0 + act
25 = 0 + 1.5t1
t1 = 16.67 s
A:
+ B v2 = v0 2 + 2ac(s - s0)
252 = 0 + 2(1.5)(s1 - 0)
s1 = 208.33 m
For stage (2) of the motion, s0 = 108.22 ft, v0 = 25 ft>s, t = 60 s, and ac = 0. Thus,
A:
+ B 1
s = s0 + v0t + a t2
2 c
s = 208.33 + 25(60) + 0
= 1708.33ft = 1708 m Ans.
The average speed of the car is then
s 1708.33
vavg = = = 22.3 m>s Ans.
t1 + t2 16.67 + 60
12–19. A car is to be hoisted by elevator to the fourth

floor of a parking garage, which is 48 ft above the ground. If
the elevator can accelerate at 0.6 ft>s2, decelerate at
0.3 ft>s2, and reach a maximum speed of 8 ft>s, determine
the shortest time to make the lift, starting from rest and
ending at rest.
+c v2 = v20 + 2 ac (s - s0)
v2max = 0 + 2(0.6)(y - 0)
0 = v2max + 2(-0.3)(48 - y)
0 = 1.2 y - 0.6(48 - y)
y = 16.0 ft, vmax = 4.382 ft>s 6 8 ft>s
+c v = v0 + ac t
4.382 = 0 + 0.6 t1
t1 = 7.303 s
0 = 4.382 - 0.3 t2
t2 = 14.61 s
t = t1 + t2 = 21.9 s Ans.
10
91962_01_s12-p0001-0176 6/8/09 8:06 AM Page 12
•12–21. Two particles A and B start from rest at the origin

s = 0 and move along a straight line such that
aA = (6t - 3) ft>s2 and aB = (12t2 - 8) ft>s2, where t is in
seconds. Determine the distance between them when
t = 4 s and the total distance each has traveled in t = 4 s.
Velocity: The velocity of particles A and B can be determined using Eq. 12-2.
dyA = aAdt
yA t
L0 L0
dyA = (6t - 3)dt
yA = 3t2 - 3t
dyB = aBdt
yB t
L0 L0
dyB = (12t2 - 8)dt
yB = 4t3 - 8t
The times when particle A stops are

3t2 - 3t = 0 t = 0 s and = 1 s
The times when particle B stops are

4t3 - 8t = 0 t = 0 s and t = 22 s
Position: The position of particles A and B can be determined using Eq. 12-1.
dsA = yAdt
sA t
L0 L0
dsA = (3t2 - 3t)dt
3 2
sA = t3 - t
2
dsB = yBdt
sB t
L0 L0
dsB = (4t3 - 8t)dt
sB = t4 - 4t2
The positions of particle A at t = 1 s and 4 s are
3 2
sA |t = 1 s = 13 - (1 ) = -0.500 ft
2
3 2
sA |t = 4 s = 43 - (4 ) = 40.0 ft
2
Particle A has traveled
dA = 2(0.5) + 40.0 = 41.0 ft Ans.
The positions of particle B at t = 22 s and 4 s are
sB |t = 12 = ( 22)4 - 4(22)2 = -4 ft
sB |t = 4 = (4)4 - 4(4)2 = 192 ft
Particle B has traveled

dB = 2(4) + 192 = 200 ft Ans.
At t = 4 s the distance beween A and B is

¢sAB = 192 - 40 = 152 ft Ans.
12
91962_01_s12-p0001-0176 6/8/09 8:06 AM Page 17
*12–28. If the effects of atmospheric resistance are

accounted for, a falling body has an acceleration defined by
the equation a = 9.81[1 - v2(10 -4)] m>s2, where v is in m>s
and the positive direction is downward. If the body is
released from rest at a very high altitude, determine (a) the
velocity when t = 5 s, and (b) the body’s terminal or
maximum attainable velocity (as t : q ).
Velocity: The velocity of the particle can be related to the time by applying Eq. 12–2.
dy
(+ T) dt =
a
t y
L0 L0 9.81[1 - (0.01y) ]
dy
dt = 2
y y
9.81 L0 2(1 + 0.01y) L0

1 dy dy
t = c + d
2(1 - 0.01y)
1 + 0.01y
9.81t = 50lna b
1 - 0.01y
100(e0.1962t - 1)
y = [1]
e0.1962t + 1
a) When t = 5 s, then, from Eq. [1]
100[e0.1962(5) - 1]
y = = 45.5 m>s Ans.
e0.1962(5) + 1
e0.1962t - 1
b) If t : q , : 1. Then, from Eq. [1]
e0.1962t + 1
ymax = 100 m>s Ans.
17
91962_01_s12-p0001-0176 6/8/09 8:25 AM Page 65
12–79. A particle travels along the path y2 = 4x with a

constant speed of v = 4 m>s. Determine the x and y
components of the particle’s velocity and acceleration when
the particle is at x = 4 m.
Velocity: The x and y components of the particle’s velocity can be related by taking
the first time derivative of the path’s equation using the chain rule.
# #
2yy = 4x
# 2 #
y = x
y
or
2
vy = v (1)
y x
At x = 4 m, y = 24(4) = 4 m. Thus Eq. (1) becomes

1
vy = v (2)
2 x
The magnitude of the particle’s velocity is
v = 2vx 2 + vy 2 (3)
Substituting v = 4 m>s and Eq. (2) into Eq. (3),
2
4 = 2 1
A vx + a v x b
2
vx = 3.578 m>s = 3.58 m>s Ans.
Substituting the result of nx into Eq. (2), we obtain
vy = 1.789 m>s = 1.79 m>s Ans.
Acceleration: The x and y components of the particle’s acceleration can be related

by taking the second time derivative of the path’s equation using the chain rule.
# #
2(yy + yy) = 4x
#
y2 + yy = 2x
or
vy 2 + yay = 2ax (4)
When x = 4 m, y = 4 m, and vy = 1.789 m>s. Thus Eq. (4) becomes
1.7892 + 4ay = 2ax
ay = 0.5ax - 0.8 (5)
Since the particle travels with a constant speed along the path, its acceleration along
the tangent of the path is equal to zero. Here, the angle that the tangent makes with the
dy 1
horizontal at x = 4 m is u = tan - 1 ¢ ≤ 2 = tan - 1 ¢ 1>2 ≤ 2 = tan - 1 (0.5) = 26.57°.
dx x = 4 m x x=4 m
Thus, from the diagram shown in Fig. a,
ax cos 26.57° + ay sin 26.57° = 0 (6)
Solving Eqs. (5) and (6) yields
ax = 0.32 m>s2 ay = -0.64 m>s2 = 0.64 m>s2 T Ans.
65
91962_01_s12-p0001-0176 6/8/09 8:26 AM Page 67
•12–81. A particle travels along the circular path from A y

to B in 1 s. If it takes 3 s for it to go from A to C, determine
its average velocity when it goes from B to C.
30! C
Position: The coordinates for points B and C are [30 sin 45°, 30 - 30 cos 45°] and 45!
[30 sin 75°, 30 - 30 cos 75°]. Thus, 30 m
rB = (30 sin 45° - 0)i + [(30 - 30 cos 45°) - 30]j

B
= {21.21i - 21.21j} m
x
A
rC = (30 sin 75° - 0)i + [(30 - 30 cos 75°) - 30]j
= {28.98i - 7.765j} m
Average Velocity: The displacement from point B to C is ¢rBC = rC - rB

= (28.98i - 7.765j) - (21.21i - 21.21j) = {7.765i + 13.45j} m.
¢rBC 7.765i + 13.45j

(vBC)avg = = = {3.88i + 6.72j} m>s Ans.
¢t 3 - 1
12–82. A car travels east 2 km for 5 minutes, then north

3 km for 8 minutes, and then west 4 km for 10 minutes.
Determine the total distance traveled and the magnitude
of displacement of the car. Also, what is the magnitude of
the average velocity and the average speed?
Total Distance Traveled and Displacement: The total distance traveled is
s = 2 + 3 + 4 = 9 km Ans.
and the magnitude of the displacement is
¢r = 2(2 + 4)2 + 32 = 6.708 km = 6.71 km Ans.
Average Velocity and Speed: The total time is ¢t = 5 + 8 + 10 = 23 min = 1380 s.

The magnitude of average velocity is
¢r 6.708 A 103 B
yavg = = = 4.86 m>s Ans.
¢t 1380
and the average speed is
s 9 A 103 B
A ysp B avg = = = 6.52 m>s Ans.
¢t 1380
67
91962_01_s12-p0001-0176 6/8/09 8:32 AM Page 79
•12–97. A boy throws a ball at O in the air with a speed v0

at an angle u1. If he then throws another ball with the same
B
speed v0 at an angle u2 6 u1, determine the time between
the throws so that the balls collide in mid air at B. u2 y
u1
O
Vertical Motion: For the first ball, the vertical component of initial velocity is
(y0)y = y0 sin u1 and the initial and final vertical positions are (s0)y = 0 and sy = y,
x
respectively.
1
(+ c ) sy = (s0)y + (y0)y t + (a ) t2
2 cy
1
y = 0 + y0 sin u1t1 + (-g)t21 [1]
2
For the second ball, the vertical component of initial velocity is (y0)y = y0 sin u2 and
the initial and final vertical positions are (s0)y = 0 and sy = y, respectively.
1
(+ c ) sy = (s0)y + (y0)y t + (a ) t2
2 cy
1
y = 0 + y0 sin u2t2 + (-g)t22 [2]
2
Horizontal Motion: For the first ball, the horizontal component of initial velocity is
(y0)x = y0 cos u1 and the initial and final horizontal positions are (s0)x = 0 and
sx = x, respectively.
A:
+ B sx = (s0)x + (y0)x t
x = 0 + y0 cos u1 t1 [3]
For the second ball, the horizontal component of initial velocity is (y0)x = y0 cos u2
and the initial and final horizontal positions are (s0)x = 0 and sx = x, respectively.
A:
+ B sx = (s0)x + (y0)x t
x = 0 + y0 cos u2 t2 [4]
Equating Eqs. [3] and [4], we have
cos u1
t2 = t [5]
cos u2 1
Equating Eqs. [1] and [2], we have
1
y0 t1 sin u1 - y0 t2 sin u2 = g A t21 - t22 B [6]
2
Solving Eq. [5] into [6] yields
2y0 cos u2 sin(u1 - u2)
t1 =
g(cos2 u2 - cos2 u1)
2y0 cos u1 sin(u1 - u2)
t2 =
g(cos2 u2 - cos2u1)
Thus, the time between the throws is
2y0 sin(u1 - u2)(cos u2 - cos u1)

¢t = t1 - t2 =
g(cos2 u2 - cos2 u1)
2y0 sin (u1 - u2)
= Ans.
g(cos u2 + cos u1)
79
91962_01_s12-p0001-0176 6/8/09 8:34 AM Page 82
*12–100. The velocity of the water jet discharging from the

orifice can be obtained from v = 22 gh, where h = 2 m is
the depth of the orifice from the free water surface. Determine 2m
A vA
the time for a particle of water leaving the orifice to reach
point B and the horizontal distance x where it hits the surface. 1.5 m
B
Coordinate System: The x–y coordinate system will be set so that its origin coincides x
with point A. The speed of the water that the jet discharges from A is
vA = 22(9.81)(2) = 6.264 m>s
x-Motion: Here, (vA)x = vA = 6.264 m>s, xA = 0, xB = x, and t = tA. Thus,
A:
+ B xB = xA + (vA)xt
x = 0 + 6.264tA (1)
y-Motion: Here, (vA)y = 0, ay = -g = -9.81 m>s2, yA = 0 m, yB = -1.5 m, and

t = tA. Thus,
1
A+cB yB = yA + (vA)yt + a t2
2 y
1
-1.5 = 0 + 0 + (-9.81)tA 2
2
tA = 0.553 s
Thus,
x = 0 + 6.264(0.553) = 3.46 m Ans.
•12–101. A projectile is fired from the platform at B. The B

shooter fires his gun from point A at an angle of 30°.
C
Determine the muzzle speed of the bullet if it hits the
projectile at C.
vA
10 m
A 30!
Coordinate System: The x–y coordinate system will be set so that its origin coincides
with point A. 1.8 m
20 m
x-Motion: Here, xA = 0 and xC = 20 m. Thus,
A:
+ B xC = xA + (vA)xt
20 = 0 + vA cos 30° t (1)
y-Motion: Here, yA = 1.8, (vA)y = vA sin 30°, and ay = -g = -9.81 m>s2. Thus,
1
A+cB yC = yA + (vA)yt + a t2
2 y
1
10 = 1.8 + vA sin 30°(t) + (-9.81)(t)2
2
Thus,
20 sin 30°
10 - 1.8 = ¢ ≤ (t) - 4.905(t)2
cos 30°(t)
t = 0.8261 s
So that
20
vA = = 28.0 m>s Ans.
cos 30°(0.8261)
82
91962_01_s12-p0001-0176 6/8/09 8:37 AM Page 85
12–106. The boy at A attempts to throw a ball over the C

roof of a barn such that it is launched at an angle uA = 40°.
Determine the minimum speed vA at which he must throw
the ball so that it reaches its maximum height at C. Also,
8m
find the distance d where the boy must stand so that he can vA
make the throw. A uA
1m
Vertical Motion: The vertical components of initial and final velocity are
d 4m
(y0)y = (yA sin 40°) m>s and yy = 0, respectively. The initial vertical position is
(s0)y = 1 m.
A+cB yy = (y0) + ac t
0 = yA sin 40° + ( -9.81) t [1]
1
A+cB sy = (s0)y + (y0)y t + (a ) t2
2 cy
1
8 = 1 + yA sin 40°t + (-9.81) t2 [2]
2
Solving Eqs. [1] and [2] yields
yA = 18.23 m>s = 18.2 m>s Ans.
t = 1.195 s
Horizontal Motion: The horizontal component of velocity is (y0)x = yA cos uA

= 18.23 cos 40° = 13.97 m>s. The initial and final horizontal positions are (s0)x = 0
and sx = (d + 4) m, respectively.
A:
+ B sx = (s0)x + (y0)x t
d + 4 = 0 + 13.97(1.195)
d = 12.7 m Ans.
12–107. The fireman wishes to direct the flow of water

A
from his hose to the fire at B. Determine two possible u
angles u1 and u2 at which this can be done. Water flows from vA
the hose at vA = 80 ft>s.
20 ft
A:
+ B s = s0 + v0 t B
35 = 0 + (80) cos u
1 2
A+cB s = s0 + v0 t + at 35 ft
2 c
1
-20 = 0 - 80 sin u t + (-32.2)t2
2
Thus,
0.4375 0.1914
20 = 80 sin u t + 16.1 ¢ ≤
cos u cos2 u
20 cos2 u = 17.5 sin 2u + 3.0816
Solving,
u1 = 25.0° (below the horizontal) Ans.
u2 = 85.2° (above the horizontal) Ans.
85
91962_01_s12-p0001-0176 6/8/09 8:39 AM Page 89
•12–113. Determine the maximum constant speed a

race car can have if the acceleration of the car cannot
exceed 7.5 m>s2 while rounding a track having a radius of
curvature of 200 m.
Acceleration: Since the speed of the race car is constant, its tangential component of
acceleration is zero, i.e., at = 0. Thus,
v2
a = an =
r
v2
7.5 =
200
n = 38.7 m>s Ans.
12–114. An automobile is traveling on a horizontal

circular curve having a radius of 800 ft. If the acceleration of
the automobile is 5 ft>s2, determine the constant speed at
which the automobile is traveling.
Acceleration: Since the automobile is traveling at a constant speed, at = 0.

y2
Thus, an = a = 5 ft>s2. Applying Eq. 12–20, an = , we have
r
y = 2ran = 2800(5) = 63.2 ft>s Ans.
12–115. A car travels along a horizontal circular curved

road that has a radius of 600 m. If the speed is uniformly
increased at a rate of 2000 km>h2, determine the magnitude
of the acceleration at the instant the speed of the car is
60 km>h.
2
2000 km 1000 m 1h
at = ¢ 2
ba ba b = 0.1543 m>s2
h 1 km 3600 s
60 km 1000 m 1h
y = a ba ba b = 16.67 m>s
h 1 km 3600 s
y2 16.672
an = = = 0.4630 m>s2
r 600
a = 2a2t + a2n = 20.15432 + 0.46302 = 0.488 m>s2 Ans.
89
91962_01_s12-p0001-0176 6/8/09 8:41 AM Page 92
*12–120. The car travels along the circular path such that
its speed is increased by at = (0.5et) m>s2, where t is in
seconds. Determine the magnitudes of its velocity and
s ! 18 m
acceleration after the car has traveled s = 18 m starting
from rest. Neglect the size of the car.
y t
L0 L0
dy = 0.5e t dt
y = 0.5(e t - 1)
18 t
L0 L0
ds = 0.5 (e t - 1)dt
ρ ! 30 m
t
18 = 0.5(e - t - 1)
Solving,
t = 3.7064 s
y = 0.5(e 3.7064 - 1) = 19.85 m>s = 19.9 m>s Ans.
#
at = y = 0.5e t ƒ t = 3.7064 s = 20.35 m>s2
y2 19.852
an = = = 13.14 m>s2
r 30
a = 2a2t + a2n = 220.352 + 13.142 = 24.2 m>s2 Ans.
•12–121. The train passes point B with a speed of 20 m>s y

which is decreasing at at = – 0.5 m>s2. Determine the x
y ! 200 e 1000
magnitude of acceleration of the train at this point.
Radius of Curvature:
x B
A
y = 200e 1000
dy x x
1 x
= 200 a be 1000 = 0.2e 1000
dx 1000 400 m
d2y 1 x x
.
2
= 0.2a be 1000 = 0.2 A 10-3 B e 1000
dx 1000
3>2
x 2
dy 2 3>2 C 1 + ¢ 0.2 e 1000 ≤ S
c1 + a b d
dx
r = = 6 = 3808.96 m
d2y x
2 2 ` 0.2 A 10 B
-3
e 1000 `
dx2
x = 400 m
Acceleration:
#
a t = v = -0.5 m>s2
v2 202
an = = = 0.1050 m>s2
r 3808.96
The magnitude of the train’s acceleration at B is
a = 2a2t + a2n = 2 A -0.5 B 2 + 0.10502 = 0.511 m>s2 Ans.
92
91962_01_s12-p0001-0176 6/8/09 8:50 AM Page 114
12–147. The box of negligible size is sliding down along a y

curved path defined by the parabola y = 0.4x2. When it is at
A (xA = 2 m, yA = 1.6 m), the speed is vB = 8 m>s and the
increase in speed is dvB>dt = 4 m>s2. Determine the
magnitude of the acceleration of the box at this instant.
A
2
y ! 0.4x
y = 0.4 x2
x
dy 2m
2 = 0.8x 2 = 1.6
dx x = 2 m x=2 m
d2y
2 = 0.8
dx2 x=2 m
C 1 + (dx ) D
dy 2 3>2
C 1 + (1.6)2 D 3>2
r = 4 = = 8.396 m
d 2y |0.8|
` 2 `
dx x=2 m
2
yB 82
an = = = 7.622 m>s2
r 8.396
a = 2a2t + a2n = 2(4)2 + (7.622)2 = 8.61 m>s2 Ans.
*12–148. A spiral transition curve is used on railroads to y

connect a straight portion of the track with a curved
portion. If the spiral is defined by the equation
y = (10 - 6)x3, where x and y are in feet, determine the
y ! (10"6)x3
magnitude of the acceleration of a train engine moving with
a constant speed of 40 ft>s when it is at point x = 600 ft.
y = (10)-6x3
dy v ! 40 ft/s
2 = 3(10)-6x2 2 = 1.08
dx x = 600 ft x = 600 ft
d2y x
2 = 6(10)-6x 2 = 3.6(10)-3
dx2 x = 600 ft x = 600 ft 600 ft
[1 + (dx ) D
dy 2 3>2
[1 + (1.08)2]3>2
r4 = 4 = = 885.7 ft
2
dy |3.6(10)-3|
` 2 `
x = 600 ft dx x = 600 ft
y2 402
an = = = 1.81 ft>s2
r 885.7
a = 2a2t + a2n = 20 + (1.81)2 = 1.81 ft>s2 Ans.
114

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91962_01_s12-p0001-0176 6/8/09 8:05 AM Page 3

•12–5. A particle is moving along a straight line with the

v = A 6t2 - 2t3>2 B ft>s Ans.

Position: Using this result and the initial condition s = 15 ft at t = 0 s,

12–6. A ball is released from the bottom of an elevator

= -90.6 ft>s = 90.6 ft>s T Ans.

12–18. A car starts from rest and moves with a constant

Kinematics: For stage (1) of the motion, v0 = 0, s0 = 0, v = 25 m>s, and

= 1708.33ft = 1708 m Ans.

The average speed of the car is then

12–19. A car is to be hoisted by elevator to the fourth

y = 16.0 ft, vmax = 4.382 ft>s 6 8 ft>s

•12–21. Two particles A and B start from rest at the origin

The times when particle A stops are

The times when particle B stops are

The positions of particle A at t = 1 s and 4 s are

The positions of particle B at t = 22 s and 4 s are

sB |t = 4 = (4)4 - 4(4)2 = 192 ft

Particle B has traveled

At t = 4 s the distance beween A and B is

*12–28. If the effects of atmospheric resistance are

9.81 L0 2(1 + 0.01y) L0

a) When t = 5 s, then, from Eq. [1]

ymax = 100 m>s Ans.

12–79. A particle travels along the path y2 = 4x with a

At x = 4 m, y = 24(4) = 4 m. Thus Eq. (1) becomes

Substituting v = 4 m>s and Eq. (2) into Eq. (3),

vx = 3.578 m>s = 3.58 m>s Ans.

Substituting the result of nx into Eq. (2), we obtain

vy = 1.789 m>s = 1.79 m>s Ans.

Acceleration: The x and y components of the particle’s acceleration can be related

When x = 4 m, y = 4 m, and vy = 1.789 m>s. Thus Eq. (4) becomes

1.7892 + 4ay = 2ax

ay = 0.5ax - 0.8 (5)

ax cos 26.57° + ay sin 26.57° = 0 (6)

Solving Eqs. (5) and (6) yields

ax = 0.32 m>s2 ay = -0.64 m>s2 = 0.64 m>s2 T Ans.

•12–81. A particle travels along the circular path from A y

rB = (30 sin 45° - 0)i + [(30 - 30 cos 45°) - 30]j

Average Velocity: The displacement from point B to C is ¢rBC = rC - rB

¢rBC 7.765i + 13.45j

12–82. A car travels east 2 km for 5 minutes, then north

Total Distance Traveled and Displacement: The total distance traveled is

and the magnitude of the displacement is

¢r = 2(2 + 4)2 + 32 = 6.708 km = 6.71 km Ans.

Average Velocity and Speed: The total time is ¢t = 5 + 8 + 10 = 23 min = 1380 s.

and the average speed is

•12–97. A boy throws a ball at O in the air with a speed v0

Equating Eqs. [3] and [4], we have

2y0 sin(u1 - u2)(cos u2 - cos u1)

*12–100. The velocity of the water jet discharging from the

vA = 22(9.81)(2) = 6.264 m>s

x-Motion: Here, (vA)x = vA = 6.264 m>s, xA = 0, xB = x, and t = tA. Thus,

y-Motion: Here, (vA)y = 0, ay = -g = -9.81 m>s2, yA = 0 m, yB = -1.5 m, and

x = 0 + 6.264(0.553) = 3.46 m Ans.

•12–101. A projectile is fired from the platform at B. The B

20 = 0 + vA cos 30° t (1)

12–106. The boy at A attempts to throw a ball over the C

0 = yA sin 40° + ( -9.81) t [1]