Solutions Manual for
Engineering Mechanics
Dynamics 16th Edition By
Russell C. Hibbeler (All
Dynamics Part Chapters 12-
22, 100% Original Verified, A+
Grade)
This is The Only Original and
Complete Solutions Manual
for Dynamics 16th Edition, All
Other Files in the Market are
Fake/Old/Wrong Edition.
,12–1. Chapter12
A baseball is thrown downward from a 50-ft tower with an
initial speed of 18 ft>s.
ft>s. Determine the speed at which it hits
the ground and the time of travel.
SOLUTION
A + T B v22 = v21 + 2ac(s2 - s1)
v22 = (18)2 + 2(32.2)(50 - 0)
v2 = 59.532 = 59.5 ft>s T Ans.
A + T B v2 = v1 + ac t
59.532 = 18 + 32.2(t)
t = 1.29 s Ans.
Ans:
v2 = 59.5 ft>s
t = 1.29 s
,12–2.
Starting from rest, a particle moving in a straight line has an
acceleration of a = (2t - 6) m>s2, where t is in seconds.
What is the particle’s velocity when t = 6 s, and what is its
position when t = 11 s?
SOLUTION
a = 2t - 6
dv = a dt
v t
L0 L0
dv = (2t - 6) dt
v = t 2 - 6t
ds = v dt
s t
L0 L0
ds = (t2 - 6t) dt
t3
s = - 3t2
3
When t = 6 s,
v = 0 Ans.
When t = 11 s,
s = 80.7 m Ans.
Ans:
v = 0
s = 80.7 m
, 12–3.
From approximately what floor of a building must a car be
dropped from an at-rest position so that it reaches a speed
of 80.7 ft>s 155
(55 mi>h2
mi>h) when it hits the ground? Each floor is
12 ft higher than the one below it. (Note: You may want to
remember this when traveling 55 mi>h.)
mi>h.)
SOLUTION
(+ T ) v 2 = v20 + 2ac(s - s0)
80.72 = 0 + 2(32.2)(s - 0)
s = 101.13 ft
101.13
# of floors = = 8.43
12
The car must be dropped from the 9th floor. Ans.
Ans:
n = 9
Engineering Mechanics
Dynamics 16th Edition By
Russell C. Hibbeler (All
Dynamics Part Chapters 12-
22, 100% Original Verified, A+
Grade)
This is The Only Original and
Complete Solutions Manual
for Dynamics 16th Edition, All
Other Files in the Market are
Fake/Old/Wrong Edition.
,12–1. Chapter12
A baseball is thrown downward from a 50-ft tower with an
initial speed of 18 ft>s.
ft>s. Determine the speed at which it hits
the ground and the time of travel.
SOLUTION
A + T B v22 = v21 + 2ac(s2 - s1)
v22 = (18)2 + 2(32.2)(50 - 0)
v2 = 59.532 = 59.5 ft>s T Ans.
A + T B v2 = v1 + ac t
59.532 = 18 + 32.2(t)
t = 1.29 s Ans.
Ans:
v2 = 59.5 ft>s
t = 1.29 s
,12–2.
Starting from rest, a particle moving in a straight line has an
acceleration of a = (2t - 6) m>s2, where t is in seconds.
What is the particle’s velocity when t = 6 s, and what is its
position when t = 11 s?
SOLUTION
a = 2t - 6
dv = a dt
v t
L0 L0
dv = (2t - 6) dt
v = t 2 - 6t
ds = v dt
s t
L0 L0
ds = (t2 - 6t) dt
t3
s = - 3t2
3
When t = 6 s,
v = 0 Ans.
When t = 11 s,
s = 80.7 m Ans.
Ans:
v = 0
s = 80.7 m
, 12–3.
From approximately what floor of a building must a car be
dropped from an at-rest position so that it reaches a speed
of 80.7 ft>s 155
(55 mi>h2
mi>h) when it hits the ground? Each floor is
12 ft higher than the one below it. (Note: You may want to
remember this when traveling 55 mi>h.)
mi>h.)
SOLUTION
(+ T ) v 2 = v20 + 2ac(s - s0)
80.72 = 0 + 2(32.2)(s - 0)
s = 101.13 ft
101.13
# of floors = = 8.43
12
The car must be dropped from the 9th floor. Ans.
Ans:
n = 9
