0320307hwkCh04
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Chapter 4: TwoDimensional Kinematics
Solutions to Problems
7.
Picture the Problem
: The arrow falls below the target center as it
flies from the bow to the target.
Strategy:
Treat the vertical and horizontal motions separately.
First
find the time required for the arrow to drop straight down 52 cm
from rest.
Then use that time together with the horizontal target
distance to find the horizontal speed of the arrow.
That must also
equal the initial speed because the arrow was launched horizontally.
Solution:
1.
Find the time to drop 52 cm:
( )
2
2 0.52 m
2
0.326 s
9.81 m/s
y
t
g
Δ
Δ =
=
=
2.
Find the speed of the arrow from the
horizontal distance and time elapsed:
0
15 m
46 m/s
0.326 s
x
x
v
t
Δ
=
=
=
Δ
Insight:
We had to bend the significant figures rules a bit to obtain an accurate answer. Another way to solve this
problem and avoid the rounding error is to solve equation 48 for velocity:
( )
2
0
2
v
g x
h
y
=

.
In this case
we would set
h
= 0.52 m ,
y
= 0.0 m, and
x
= 15 m.

8.
Picture the Problem
: The water falls down along a parabolic arc, maintaining its horizontal velocity but gaining vertical
speed as it falls.
Strategy
: Find the vertical speed of the water after falling 108 m.
The horizontal velocity remains constant throughout
the fall.
Then find the magnitude of the velocity from the horizontal and vertical components.
Solution:
1.
Use equations 46 to find
y
v
:
( )( )
2
2
2
2
2
0
2
0
2 9.81 m/s
0 m –108 m
2120 m /s
y
y
v
v
g y
=

Δ =

=
2.
Use the components of the velocity to find
the speed:
( )
2
2
2
2
2
3.60 m/s
2120 m /s
46.2 m s
x
y
v
v
v
=
+
=
+
=
Insight:
Projectile problems are often solved by first considering the vertical motion, which determines the time of flight
and the vertical speed, and then considering the horizontal motion.

16.
Picture the Problem
: The pumpkin’s trajectory is depicted in the figure at right.
Strategy:
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 Spring '08
 CROFT
 Physics, Velocity, m/s

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