Posted by tanya on Wednesday, December 2, 2009 at 7:57pm.
Before trying to answer (a), let's do (b).
Due to the starting kinetic energy, the KE at the bottom will be increased from
(1/2)M(7.5)^2 to (1/2)M[(1.5)^2 + (7.5)^2] = (1/2)M(58.5) = (1/2)M(7.64)^2
The velocity at the bottom in the second case will be 7.64 m/s.
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