SOLUTION
If we assume that upward is the positive direction, the initial speed of the
pellet is, from Equation 2.9,
2
2
2
0
2
( – 27 m/s)
2(–9.80 m/s
)(– 15 m)
20.9 m/s
v
v
ay
=

=

=
Chapter 2
Problems
73
Equation 2.9 can again be used to find the maximum height of the pellet if it were fired
straight up.
At its maximum height,
v
= 0 m/s, and Equation 2.9 gives
2
2
0
2
(20.9 m/s)
22 m
2
2(–9.80 m/s
)
v
y
a


=
=
=
_____________________________________________________________________________
_
53.
SSM
REASONING AND SOLUTION
Since the balloon is released from rest, its initial
velocity is zero.
The time required to fall through a vertical displacement
y
can be found
from Equation 2.8
(
29
2
1
0
2
y
v t
at
=
+
with
0
0
v
=
m/s.
Assuming upward to be the positive
direction, we find
2
2
2(–6.0 m)
1.1 s
–9.80 m/s
y
t
a
=
=
=
_____________________________________________________________________________
_
54.
REASONING
The initial speed of the ball can be determined from Equation 2.9
(
29
2
2
0
2
v
v
ay
=
+
. Once the initial speed of the ball is known, Equation 2.9 can be used a
second time to determine the height above the launch point when the speed of the ball has
decreased to one half of its initial value.
SOLUTION
When the ball has reached its maximum height, its velocity is zero.
If we
take upward as the positive direction, we have from Equation 2.9
(
29
2
2
2
0
2
0 m/s
2(–9.80 m/s
)(12.0 m)
15.3 m/s
v
v
ay
=

=

= +
When the speed of the ball has decreased to one half of its initial value,
1
0
2
v
v
=
, and
Equation 2.9 gives
2
2
2
2
2
1
2
0
0
0
0
2
2
(
)
1
(+15.3 m/s)
1
1
1
8.96 m
2
2
2
4
4
2(–9.80 m/s
)
v
v
v
v
v
y
a
a
a


=
=
=

=

=
÷
÷
55.
REASONING
The displacement
y
of the diver is equal to her average velocity
v
multiplied by the time
t
, or
y
vt
=
. Since the diver has a constant acceleration (the
acceleration due to gravity), her average velocity is equal to
(
29
1
0
2
v
v
v
=
+
,
where
v
0
and
v
are, respectively, the initial and final velocities. Thus, according to Equation 2.7, the
displacement of the diver is
(
29
1
0
2
y
v
v t
=
+
(2.7)
74
KINEMATICS IN ONE DIMENSION
The final velocity and the time in this expression are known, but the initial velocity is not.
To determine her velocity at the beginning of the 1.20s period (her initial velocity), we turn
to her acceleration. The acceleration is defined by Equation 2.4 as the change in her
velocity,
v

v
0
, divided by the elapsed time
t
:
(
29
0
/
a
v
v
t
=

. Solving this equation for the
initial velocity
v
0
yields
0
v
v
at
=

Substituting this relation for
v
0
into Equation 2.7, we obtain
(
29
(
29
2
1
1
1
0
2
2
2
y
v
v t
v
at
v t
vt
at
=
+
=

+
=

SOLUTION
The diver’s acceleration is that due to gravity, or
a
=

9.80 m/s
2
. The
acceleration is negative because it points downward, and this direction is the negative
direction. The displacement of the diver during the last 1.20 s of the dive is
(
29
(
29
(
29
(
29
2
2
2
1
1
2
2
10.1 m/s
1.20 s
9.80 m/s
1.20 s
5.06 m
y
vt
at
=

= 


= 
The displacement of the diver is negative because she is moving downward.
_____________________________________________________________________________
_
56.
REASONING
The ball is initially in free fall, then collides with the pavement and
rebounds, which puts it into free fall again, until caught by the boy.
We don’t have enough
information to analyze its collision with the pavement, but we’re only asked to calculate the
time it spends in the air, undergoing freefall motion.
The motion can be conveniently
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 Spring '11
 rollino
 Physics, Velocity, m/s