Posted by Bartholomew on Thursday, January 15, 2009 at 7:57pm.
Have you tried applying conservation of energy? Show your work and one of us will gladly critique it.
(V2^2)=(Vi)^2+2ad
0=16-19.6d
-16=-19.6d
d=0.816 m + 30 m = 30.816 meters above the ground
mgh = mgh + 1/2kx^2 = (5 kg)(9.8 m/s^2)(30.816) = 200x^2-49x+490
x=2.75 m
mgh=1/2kx^2+mgh+1/2mv^2
(5)(9.8)(30.186)=1/2(400)(0.3)^2+(5)(9.8)(9.7)+1/2(5v^2)
v=20.17 m/s
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