fullwood (tf5349) – Homework #4 Projectile – Catala – (49033)
1
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printout
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have
22
questions.
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before answering.
001
10.0 points
A fireman, 56
.
9 m away from a burning build
ing,
directs a stream jet of water from a
ground level fire hose at an angle of 30
.
8
◦
above the horizontal.
The acceleration of gravity is 9
.
8 m
/
s
2
.
If the speed of the stream as it leaves the
hose is 40
.
2 m
/
s, at what height will the
stream of water strike the building?
Correct answer: 20
.
614 m.
Explanation:
56
.
9 m
40
.
2 m
/
s
30
.
8
◦
Δ
y
Given :
d
= 56
.
9 m
,
v
i
= 40
.
2 m
/
s
,
and
θ
= 30
.
8
◦
.
The components of the initial velocity are
v
ix
=
v
i
cos
θ
,
and
v
iy
=
v
i
sin
θ .
The horizontal motion carries the water to
the building, so
Δ
x
=
v
ix
t
= (
v
i
cos
θ
)
t
t
=
Δ
x
v
i
cos
θ
=
56
.
9 m
(40
.
2 m
/
s)(cos 30
.
8
◦
)
= 1
.
64783 s
.
The height the water reaches at this time is
Δ
y
=
v
iy
t
−
1
2
g t
2
= (
v
i
sin
θ
)
t
−
1
2
g t
2
= (40
.
2 m
/
s)(sin 30
.
8
◦
) (1
.
64783 s)
−
1
2
(
9
.
8 m
/
s
2
)
(1
.
64783 s)
2
=
20
.
614 m
.
002
10.0 points
A daredevil decides to jump a canyon.
Its
walls are equally high and 10 m apart.
He
takes off by driving a motorcycle up a short
ramp sloped at an angle of 14
◦
.
The acceleration of gravity is 9
.
8 m
/
s
2
.
What minimum speed must he have in or
der to clear the canyon?
Correct answer: 14
.
448 m
/
s.
Explanation:
14
◦
10 m
Note:
Figure is not drawn to scale
Given :
Δ
x
= 10 m
,
θ
= 14
◦
,
and
g
= 9
.
8 m
/
s
2
.
We can find the time of flight from the hori
zontal motion:
Δ
x
=
v
i
(cos
θ
) Δ
t
Δ
t
=
Δ
x
v
i
(cos
θ
)
.
From the vertical motion,
Δ
y
=
v
i
(sin
θ
) Δ
t
−
1
2
g
(Δ
t
)
2
= 0
v
i
(sin
θ
)
−
1
2
g
Δ
t
= 0
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fullwood (tf5349) – Homework #4 Projectile – Catala – (49033)
2
so that
v
i
(sin
θ
) =
1
2
g
bracketleftbigg
Δ
x
v
i
(cos
θ
)
bracketrightbigg
v
2
i
=
g
Δ
x
2 (sin
θ
) (cos
θ
)
v
i
=
radicalBigg
g
Δ
x
2 (sin
θ
) (cos
θ
)
=
radicalBigg
(9
.
8 m
/
s
2
) (10 m)
2 (cos 14
◦
) (sin 14
◦
)
=
14
.
448 m
/
s
.
003
10.0 points
Cliff divers at Acapulco jump into the sea
from a cliff 37
.
4 m high.
At the level of the
sea, a rock sticks out a horizontal distance of
14
.
79 m.
The acceleration of gravity is 9
.
8 m
/
s
2
.
With what minimum horizontal velocity
must the cliff divers leave the top of the cliff if
they are to miss the rock?
Correct answer: 5
.
35341 m
/
s.
Explanation:
Given :
Δ
h
=
−
37
.
4 m
and
Δ
x
= 14
.
79 m
.
The time of flight is defined by the vertical
drop:
h
=
v
iy
t
−
1
2
g t
2
=
−
1
2
g t
2
since
v
iy
= 0 m/s, so
t
=
radicalBigg
−
2
h
g
The horizontal velocity is constant, so
Δ
x
=
v
x
Δ
t
v
x
=
Δ
x
Δ
t
= Δ
x
radicalbigg
−
g
2 Δ
h
= 14
.
79 m
radicalBigg
−
9
.
8 m
/
s
2
2 (
−
37
.
4 m)
=
5
.
35341 m
/
s
.
004
10.0 points
A ball is thrown and follows the parabolic
path shown. Air friction is negligible. Point
Q
is the highest point on the path. Points
P
and
R
are the same height above the ground.
Q
R
P
Which of the following diagrams best indi
cates the direction of the acceleration, if any,
on the ball at point
P
?
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 Spring '10
 PRO.KIROSHIMA
 Acceleration, Work, Velocity, Correct Answer, Standard gravity

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