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Using Equation 426, the range of the athlete without using halteres is
2
2
00
0
2
sin 2
(10.31m/s) sin 2(22.8 )
7.75 m.
9.8 m/s
v
R
g
θ
°
==
=
On the other hand, if two halteres of mass
m
= 5.50 kg were thrown at the maximum
height, then, by momentum conservation, the subsequent speed of the athlete would be
2
(2
)
x
xx
x
Mm
M
mv
Mv
v
v
M
+
′′
+=
¡
=
Thus, the change in the
x
component of the velocity is
0
0
2
2
2(5.5 kg)
(9.5 m/s) 1.34 m/s.
78 kg
xxx
x
x
x
m
vvv
v
v
v
MM
+
′
Δ
=
−=
=
=
The maximum height is attained when
0
0
yy
vv
g
t
=−
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics

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