Posted by gaby on Thursday, December 4, 2008 at 9:03pm.
Use a conservation of energy approach for both problems. The CHANGE in (1/2)M V^2 equals the change in M g H.
That means the change in V^2 equals 2 g times the change in height above the water, H.
For the second problem, the water is 10 m below but he had a speed of V^2 = 2 m/s at the start.
Vfinal^2 = Vinitial^2 + 2 g h
= 4.0 + 2*9.8*10 = 200
V = 10 m/s
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