A baseball outfielder throws a .2 kg baseball at a speed of 50 m/s and initial angle of 30 degrees

what is the kinetic energy of the ball right before it hits the ground?
what is the kinetic energy at the highest point?
what is the potential energy at the highest point?
what is its total time in the air?

for the kinetic energy at the highest point i got 50cos30= 43.3 m/s and then .5mv^2 and that equals 187.5 joules.

a) KEғɪɴaʟ = KEiɴɪтɪaʟ

      KEi = (½) • m • (Vi)²
      KEi = (½) • (0.2) • (50)²
      KEi = 250 Joules = KEғ

b) KEp = KE at peak height where [Vyp = 0]

   KEp = (½) • m • (Vp)²
   KEp = (½) • m • (Vx)²
   KEp = (½) • (0.2) • [ 50 • cos(30) ]²
   KEp = 187.5 Joules  ...   you're correct

c) PEp = m • g • (Hмax) = KEi − KEp ... no need to determine Hмax

         PEp = KEi − KEp
         PEp = 250 − 187.5
         PEp = 62.5 Joules

d) As mentioned before [Vyp = 0] so the rise time (to the peak) is:

  Vyp = g • t + Vyi
     0 = (-9.8) • t +  [ 50 • sin(30) ]
      t = 2.55 sec ... time to peak

Since the rise time = fall time,

the Time of Flight = 2 • (2.55) = 5.1 sec