This preview shows page 1. Sign up to view the full content.
2
0
2
2(9.8 m/s )(0.76 m)
3.86 m/s .
vg
y
==
=
This is a consequence of Eq. 216 where velocity
v
vanishes. As the player reaches
y
1
=
0.76 m – 0.15 m = 0.61 m, his speed
v
1
satisfies
vv
g
y
0
2
1
2
1
2
−=
, which yields
22
2
10
1
2
(3.86 m/s)
2(9.80 m/s )(0.61 m)
1.71 m/s .
g
y
=−
=
−
=
The time
t
1
that the player spends
ascending
in the top
Δ
y
1
= 0.15 m of the jump can now
be found from Eq. 217:
()
11
1
1
2 0.15 m
1
0.175 s
2
1.71m/s
0
yv
v
t
t
Δ=
+
¡
+
which means that the total time spent in that top 15 cm (both ascending and descending)
is 2(0.175 s) = 0.35 s = 350 ms.
(b) The time
t
2
when the player reaches a height of 0.15 m is found from Eq. 215:
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '06
 Stoler
 Physics

Click to edit the document details