Rearranging terms, the solutions for
f
and
τ
in terms of
V
,
g
and
θ
are as follows.
f
=
2
V
2
tan
θ
g
cos
θ
and
τ
=
2
V
tan
θ
g
(b)
The ski jumper’s angle relative to the horizontal is simply the arctangent of the ratio of the jumper’s
vertical to horizontal velocity components,

˙
z/
˙
x

, wherefore
φ
=tan
−
1
p
g
τ
V
Q
−
θ
where we subtract the slope angle,
θ
,sotha
t
φ
is the jumper’s angle relative to the slope. From the solution
to Part (a), we have
g
τ
V
=
2
V
tan
θ
V
=2tan
θ
Therefore, the ski jumper’s angle relative to the slope at the moment the jumper lands is
φ
=tan
−
1
(2 tan
θ
)
−
θ
(c)
In order to find the maximum value of
φ
, we differentiate the result of Part (b) with respect to
θ
, i.e.,
d
φ
d
θ
=
1
1+4tan
2
θ
d
d
θ
(2 tan
θ
)
−
1=
2sec
2
θ
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 Spring '06
 Shiflett
 Calculus, Derivative, Slope, Ratio, Inverse trigonometric functions, ski jumper

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