(b) The height of the person when he is directly above the second wheel can be found by
solving Eq. 424. With the second wheel located at
23 m (23/ 2) m
34.5 m,
x
=+
=
we
have
22
2
00
2
2
(9.8 m/s )(34.5 m)
tan
3.0 m (34.5 m)tan53
2
cos
2(26.52 m/s) (cos53 )
25.9 m.
gx
yy x
v
θ
−
=
+
°
−
°
=
Therefore, the clearance over the second wheel is
25.9 m 18 m
7.9 m
w
yyh
Δ= − =
−
=
.
(c) The location of the center of the net is given by
2
2
sin 2
(26.52 m/s) sin(2 53 )
0t
a
n
6
9
m
.
2
cos
9.8 m/s
v
gx
x
vg
⋅°
=− =
−
¡
==
=
44. (a) Using the fact that the person (as the projectile) reaches the maximum height over
the middle wheel located at
23 m (23/ 2) m
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 Spring '08
 Any
 Physics

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