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Oslund (hno56) – Homework 2 – sai – (58015)
1
This printout should have 27 questions.
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beFore answering.
001
10.0 points
A ball is thrown upward.
Its initial verti
cal speed is 11
.
1m
/
s
,
acceleration oF gravity
is 9
.
8m
/
s
2
,
and maximum height
h
max
are
shown in the fgure below.
Neglect:
Air resistance.
11
.
1m
/
s
9
8m
2
h
max
What is its maximum height,
h
max
?
Correct answer: 6
.
28622 m.
Explanation:
Let :
v
0
= 11
.
1m
/
s
and
g
=9
.
8m
/
s
2
.
Basic Concept:
±or constant accelera
tion, we have
v
2
=
v
2
0
+2
a
(
y

y
0
)
.
(1)
Solution:
The velocity at the top is zero.
Since we know velocities and acceleration, Eq.
1 containing
v
,
a
,
y
0
, and
y
(no time) is
the easiest one to use. Choose the positive
direction to be up; then
a
=

g
and
0=
v
2
0
+ 2 (

g
)(
h
max

0)
or
h
max
=
v
2
0
2
g
=
(11
.
1m
/
s)
2
2 (9
.
8m
/
s
2
)
=6
.
28622 m
.
002
(part 1 oF 2) 10.0 points
A ball has an initial speed oF 19 m
/
s.
The acceleration oF gravity is 9
.
8m
/
s
2
.
t
h
y
2
.
9s
19 m
/
s
What will be its position aFter 2
.
9 s iF it is
thrown
a) down with an initial speed oF 19 m
/
s?
Correct answer:

96
.
309 m.
Explanation:
Basic Concepts:
This problem involves
initial velocities directed both downward and
upward, so all downward motion must be neg
ative. Assuming the initial position is
h
o
= 0,
the position at any time t is given by
h
f
=
v
o
t

1
2
gt
2
.
Let :
t
=2
.
9s
,
v
o
= 19 m
/
s
,
and
g
=9
.
8m
/
s
2
.
Solution:
h
1
=

vt

1
2
gt
2
=

(19 m
/
s) (2
.
9 s)

1
2
(9
.
8m
/
s
2
) (2
.
9 s)
2
=

96
.
309 m
.
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View Full DocumentOslund (hno56) – Homework 2 – sai – (58015)
2
003
(part 2 of 2) 10.0 points
t
h
y
2
.
9s
19 m
/
s
b) up with an initial speed of 19 m
/
s?
Correct answer: 13
.
891 m.
Explanation:
h
2
=
vt

1
2
gt
2
= (19 m
/
s) (2
.
9 s)

1
2
(9
.
8m
/
s
2
) (2
.
9 s)
2
=
13
.
891 m
.
004
10.0 points
You and your friend throw balloons Flled with
water from the roof of a several story apart
ment house. You simply drop a balloon from
rest. A second balloon is thrown downward
by your friend 3
.
7 s later with an initial speed
of 72
.
52 m
/
s. They hit the ground simultane
ously.
The acceleration of gravity is 9
.
8m
/
s
2
.
You
can neglect air resistance.
How high is the apartment house?
Correct answer: 150
.
932 m.
Explanation:
±or balloon 1:
y
1
=
1
2
gt
2
1
±or balloon 2:
y
2
=
v
0
t
2
+
gt
2
2
2
Balloon 2 is thrown Δ
t
seconds after balloon
1 and they hit the ground at the same time,
so:
t
1
=
t
2
+ Δ
t
Since the height of the house is the same for
both of the balloons,
y
1
=
y
2
gt
2
1
2
=
v
0
(
t
1

Δ
t
)+
g
2
(
t
1

Δ
t
)
2
(
v
0

g
Δ
t
)
t
1
=
v
0
Δ
t

g
2
Δ
t
2
t
1
=
v
0
Δ
t

g
2
Δ
t
2
v
0

g
Δ
t
=
Δ
t
2
+
v
0
v
0

g
Δ
t
Δ
t
2
±or Δ
t
= 2 s
t
1
= 1 +
v
0
v
0

2
g
±or
v
0
= 30 m/s, Δ
t
= 2 s
t
1
=5
.
55 s
±rom the above, once
t
1
is known, the height
of the house is given simply by
y
=
gt
2
1
2
±or
t
1
=5
.
55 s,
y
= 150
.
932 m
005
10.0 points
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 Spring '08
 Kaplunovsky
 Physics, Work

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