Oregon State University
Physics 201
,
Fall 2009
HW3
W
(due Oct. 19 at 5:00 p.m.)
Page
1
Oregon State University
Physics 201
Fall Term,
l
2008
HW3 Solutions
1.
While installing an antenna on your roof, you lose your footing and slide off the roof.
Fortu
nately, you have placed a safety net whose center is positioned at a horizontal distance of 4.15
m, measured from a point directly below the edge of the roof—and indeed you land in the cen
ter of the net.
If the roof slopes downward at a 26° angle below the horizontal, and your speed
as you leave the roof is 3.70 m/s, how high is the edge of the roof above the level of the net?
In this case, there is enough information to calculate a numerical value
with a single equation
:
You
can use the
x
motion information (i.e. the
x
displacement) to calculate the time interval of the fall,
then use that
value to
calculate how far the drop was from roof to net.
Notice that if you set your
tape measure at the point where you leave the roof, your final position when hitting the net will be
given by (
Δ
x
,
Δ
y
).
Notice, too, that
an angle measured below the horizontal is negative.
roof
v
i.x
Δ
x
θ
i
v
i.y
v
i
What you know
:
x
y
Δ
x
=
4.15
Δ
y
=
?
v
i.x
=
x
v
i
cos
θ
i
v
i.y
=
v
i
sin
θ
i
v
f
.x
=
v
i
cos
θ
i
v
f
.
y
=
?
a
x
=
0
a
y
=
–9.80
Δ
y
Δ
t
=
??
And:
θ
i
= –26°
v
i
= 3.70 m/s
I.
Calculate the time for the fall:
Use this equation
:
Δ
x
=
v
i.x
(
Δ
t
) + (
1
/
2
)
a
x
(
Δ
t
)
2
Use these knowns:
Δ
x
=
4.15
Solve for:
Δ
t
(
1.2479 s
)
v
i.x
=
x
3.70cos(–26°)
a
x
=
0
II.
Calculate the vertical displacement during the fall:
Use this equation
:
Δ
y
=
v
i.y
(
Δ
t
) + (
1
/
2
)
a
y
(
Δ
t
)
2
Use these knowns:
v
i.y
=
3.70sin(–26°)
Solve for:
Δ
y
(–
9.65 m
)
Δ
t
=
1.2479 (from
part I
)
a
y
=
–9.80
The safety
net was 9.65 m below the edge of the roof.
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Oregon State University
Physics 201
,
Fall 2009
HW3
W
(due Oct. 19 at 5:00 p.m.)
Page
2
2.
A golfer hits a golf ball eastward from an elevated tee area onto a level fairway below.
1.35 sec
onds after the ball is hit, it reaches its maximum height.
The initial angle of the ballʼs motion
off the tee is 24.0° above the horizontal.
If the tee level is 12.0 m above the level of the fairway,
how far east of the tee does the ball hit the fairway?
Δ
y
1
v
i
v
i.y
θ
i
v
i.x
x
Δ
y
2
Δ
x
1
Δ
x
2
For the trip from tee to peak height,
Then, for the trip from tee to
impact,
what you know
:
what you know
:
x
y
x
y
Δ
x
1
=
?
Δ
y
1
=
?
Δ
x
2
=
(
step IV
)
Δ
y
2
=
–12.0
v
i.x
=
x
(
step II
)
v
i.y
=
(
step I
)
v
i.x
=
x
(
step II
)
v
i.y
=
(
step I
)
v
f
.x
=
(
step II
)
v
f
.
y
=
0
v
f
.x
=
(
step II
)
v
f
.
y
=
?
a
x
=
0
a
y
=
–9.80
a
x
=
0
a
y
=
–9.80
Δ
t
1
=
1.35
Δ
t
2
=
(
step III
)
And:
θ
i
= 24°
And:
θ
i
= 24°
I.
Calculate the
initial
y
velocity of the ball as it leaves the tee:
Use this equation
:
v
f
.
y
=
v
i.y
+
a
y
(
Δ
t
)
Use these knowns:
v
f
.
y
=
0
(peak trajectory is horizontal)
Solve for:
v
i.y
(
13.23 m/s
)
a
y
=
–9.80 (known)
Δ
t
1
=
1.35
(given)
II.
Calculate the
initial
x
velocity of the ball as it leaves the tee:
Use this equation
:
tan(
θ
i
)
=
v
i.y
/
v
i.x
Use these knowns:
θ
i
=
24
(given)
Solve for:
v
i.x
(
x
29.715 m/s
)
v
i.y
=
13.23 (from
part I
)
III.
Calculate the
time from tee to impact:
Use this equation
:
Δ
y
2
=
v
i.y
(
Δ
t
2
) + (
1
/
2
)
a
y
(
Δ
t
2
)
2
Use these knowns:
Δ
y
2
=
–12.0 (given)
Solve for:
Δ
t
2
(
3.4168 s
)
v
i.y
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