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which gives
sin
(7.27 m)sin(11.3 )
1.42 m.
yd
α
=−
° =−
Therefore, at landing the skier is approximately 1.4 m below the launch level.
(c) The time it takes for the skier to land is
00
cos
(7.27 m)cos(11.3 )
0.72 s
cos
(10 m/s)cos(9.0 )
x
xd
t
vv
θ
°
==
=
=
°
.
Using Eq. 423, the
x
and
y
components of the velocity at landing are
2
cos
(10 m/s)cos(9.0 ) 9.9 m/s
sin
(10m/s)sin(9.0 ) (9.8m/s )(0.72s)
5.5m/s
x
y
g
t
°
=
=
°
−
=
−
Thus, the direction of travel at landing is
11
tan
29.1 .
9.9m/s
y
x
v
v
−−
§·
−
=
−
°
¨¸
©¹
or 29.1
°
below the horizontal. The result implies that the angle between the skier’s path
and the slope is
29.1
11.3
17.8
φ
=°
−
°
, or approximately 18
°
to two significant figures.
49. (a) The skier jumps up at an angle of
0
9.0
up from the horizontal and thus
returns to the launch level with his velocity vector 9.0
°
below the horizontal. With the
snow surface making an angle of
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 Spring '08
 Any
 Physics

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