le (txl65) – 2D Kinematics II – mohanty – (81440)
1
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001
10.0points
A plane drops a hamper of medical supplies
from a height of 2690 m during a practice run
over the ocean. The plane’s horizontal veloc
ity was 131 m
/
s at the instant the hamper
was dropped.
The acceleration of gravity is 9
.
8 m
/
s
2
.
What is the magnitude of the overall veloc
ity of the hamper at the instant it strikes the
surface of the ocean?
Correct answer: 264
.
358 m
/
s.
Explanation:
BasicConcept:
Motion in gravity field
v
2
= 2
g h
Solution:
This is a projectile motion problem.
The
motion of the dropping hamper consists of two
parts: horizontally, it moves with the initial
velocity of the plane, i.e.
v
h
=
v
= 131 m
/
s ;
vertically, due to Gravity, it moves as a freely
falling body.
Applying the equation above
gives the vertical velocity as
v
v
=
radicalbig
2
g h
=
radicalBig
2 (9
.
8 m
/
s
2
) (2690 m)
= 229
.
617 m
/
s
Thus the overall velocity at the instant the
hamper strikes the surface of the ocean is
v
f
=
radicalBig
v
2
v
+
v
2
h
=
radicalBig
(229
.
617 m
/
s )
2
+ (131 m
/
s )
2
= 264
.
358 m
/
s
002
10.0points
A toy cannon fires a 0
.
099 kg shell with initial
velocity
v
i
= 8
.
8 m
/
s in the direction
θ
= 46
◦
above the horizontal.
Δ
x
Δ
h
8
.
8 m
/
s
46
◦
Δ
y
y
The shell’s trajectory curves downward be
cause of gravity, so at the time
t
= 0
.
549 s
the shell is below the straight line by some
vertical distance Δ
h
.
Find this distance Δ
h
in the absence of
air resistance.
The acceleration of gravity is
9
.
8 m
/
s
2
.
Correct answer: 1
.
47686 m.
Explanation:
In the absence of gravity, the shell would fly
along the straight line at constant velocity:
ˆ
x
=
t v
i
cos
θ ,
ˆ
y
=
t v
i
sin
θ .
The gravity does not affect the
x
coordinate
of the shell, but it does pull its
y
coordinate
at constant downward acceleration
a
y
=

g
,
hence
x
=
t v
i
cos
θ,
y
=
t v
i
sin
θ

g t
2
2
.
Thus,
x
= ˆ
x
but
y
= ˆ
y

1
2
gt
2
, or in other
words, the shell deviates from the straightline
path by the vertical distance
Δ
h
=

ˆ
y

y

=
g t
2
2
.
Note:
This result is completely indepen
dent on the initial velocity
v
i
or angle
θ
of
the shell. It is a simple function of the flight
time
t
and nothing else (besides the constant
g
= 9
.
8 m
/
s
2
).
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le (txl65) – 2D Kinematics II – mohanty – (81440)
2
Δ
h
=
g t
2
2
=
(9
.
8 m
/
s
2
) (0
.
549 s)
2
2
=
1
.
47686 m
.
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 Spring '11
 sdsd
 Acceleration, General Relativity, Velocity, Correct Answer

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