alexander (jra2623) – oldhomework 07 – Turner – (92510)
1
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001
(part 1 of 2) 10.0 points
You are standing at the top of a cliff that has
a stairstep configuration. There is a vertical
drop of 6 m at your feet, then a horizontal
shelf of 10 m
,
then another drop of 4 m to the
bottom of the canyon, which has a horizontal
floor.
You kick a 0
.
31 kg rock, giving it an
initial horizontal velocity that barely clears
the shelf below.
10 m
Δ
x
10 m
v
6 m
4 m
What initial horizontal velocity
v
will be
required to barely clear the edge of the shelf
below you?
The acceleration of gravity is
9
.
8 m
/
s
2
.
Consider air friction to be negligi
ble.
Correct answer: 9
.
03696 m
/
s.
Explanation:
Let :
y
1
= 6 m
,
y
2
= 4 m
,
and
x
= 10 m
.
Since all of the vertical motion is down,
we can consider down to be positive, and
deal with positive gravity, vertical velocities,
and distances.
Thus
v
oy
= 0, and vertical
distances are defined by
y
=
1
2
g t
2
There is an initial horizontal velocity but
a
ox
= 0, so horizontal distances are defined
by
x
=
v
ox
t
Once launched, the vertical motion defines
the time to drop to the first shelf
2
y
1
=
g t
2
1
so
t
1
=
radicalbigg
2
y
1
g
Thus the horizontal distance traveled is
x
=
v
ox
t
1
so
v
ox
=
x
t
1
=
x
radicalbigg
g
2
y
1
= (10 m)
radicalBigg
(9
.
8 m
/
s
2
)
2 (6 m)
=
9
.
03696 m
/
s
.
002
(part 2 of 2) 10.0 points
How far from the bottom of the second cliff
will the projectile land?
Correct answer: 2
.
90994 m.
Explanation:
The time to drop the distance
y
=
y
1
+
y
2
to the bottom of the cliff is
t
=
radicalBigg
2 [
y
1
+
y
2
]
g
,
and the total horizontal distance traveled in
that time is
x
2
=
v
ox
t
=
v
ox
radicalBigg
2 [
y
1
+
y
2
]
g
= (9
.
03696 m
/
s)
radicalBigg
2 [(6 m) + (4 m)]
9
.
8 m
/
s
2
= 12
.
9099 m
.
Thus the projectile will land a distance
Δ
x
=
x
2

x
= (12
.
9099 m)

(10 m)
=
2
.
90994 m
,
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 Summer '08
 Kaplunovsky
 Work, Velocity, Correct Answer, m/s, Alexander

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