A diver of mass m drops from a board 12.0 m above the water's surface, as shown in the figure. Neglect air resistance. Suppose the diver vaults off the springboard, leaving it with an initial speed of 3.96 m/s upward. Use energy conservation to find his speed when he strikes the water.
Formula:(1/2)(m)(Vi)+(m)(g)(Yi)=(1/2)(Vf^2)+(m)(g)(Yf)
My work: Vf=sqrt(2((1/2)(3.96)+(9.8)(3.96)-12)=15.1