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58. To find the “launch” velocity of the rock, we apply Eq. 211 to the maximum height
(where the speed is momentarily zero)
()
2
00
0
9.8 m/s
2.5 s
vv g
t
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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This note was uploaded on 09/03/2009 for the course PHYS 1200 taught by Professor Stoler during the Spring '06 term at Rensselaer Polytechnic Institute.
 Spring '06
 Stoler
 Physics

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