58. To find the “launch” velocity of the rock, we apply Eq. 211 to the maximum height
(where the speed is momentarily zero)
(
)
(
)
2
0
0
0
9.8 m/s
2.5 s
v
v
gt
v
=
−
¡
=
−
so that
v
0
= 24.5 m/s (with +
y
up). Now we use Eq. 215 to find the height of the tower
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 Spring '06
 Stoler
 Physics, m/s, The Rock

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