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Unformatted text preview: 77. To get the time to reach the maximum height, we use Eq. 4-23, setting the left-hand side to zero. Thus, we find (60 m/s)sin(20o) = 2.094 s. t= 9.8 m/s2 Then (assuming α = 0) Eq. 10-13 gives θ − θo = ωo t = (90 rad/s)(2.094 s) = 188 rad,
which is equivalent to roughly 30 rev. ...
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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