Potential and Kinetic Energy Problem-high school physics?

Given:
Fa = 15.3 N upward
Fg = 11.7 N downward

Find:
a)  K.E. = ?  at t = 6.53 sec and  Vi = 0
b)  delta P.E. = increase in potential energy

Solution:
a)  K.E. = (1/2)mv^2
     K.E. = (1/2)(Fg/g)v^2
     K.E. = (1/2)(11.7/9.8)v^2
Solving for v, final velocity:
     Fnet = ma
     Fa - Fg = (Fg/g)(a)
     a = ((Fa - Fg)/(Fg/g)
     a = (15.3 - 11.7)/(11.7/9.8)
     a = 4.3 m/s^2

     a = (v - vi)/t
     v = vi + at
     v = 0 + (4.3m/s^2)6.53s
     v = 28.1 m/s
Hence,
K.E. = (1/2)(11.7/9.8)v^2
K.E. = (1/2)(11.7/9.8)(28.1)^2
K.E. = 471 joules   ANS.

b)  delta P.E.g = mgh
     delta P.E.g = (11.7)(h)

Solving for h:
         2ah = v^2 - vi^2
         h = (v^2 - vi^2)/2a
         h = (28.1^2 - 0)/2(4.3)
         h = 91.8 m
      delta P.E.g = (11.7)(91.8)
      delta P.E.g = 1074.8 joules


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