Abaseball outfielder throws a 0.150-kg baseball at a speed of 40.8 m/s and an initial angle of 28.0°. what is the kinetic energy of the baseball at the highest point of its trajectory?

Assume that aerodynamic resistance is ignored in the analysis.

The launch velocity is 40.8 m/s at an initial angle of 28°.
Therefore:
Vertical component of launch velocity is
  v = (40.8 m/s)*sin(28) = 19.154 m/s

Horizontal component of launch velocity is 
  u = (40.8 m/s)*cos(28) = 36.024 m/s

At maximum height, the vertical component of velocity is zero, but the horizontal component remains the same because aerodynamic resistance is ignored.

Given: m = 0.15 kg, mass of baseball

The kinetic energy at maximum height is
KE = (1/2)*m*u²
      = (1/2)*(0.15 kg)*(36.024 m/s)^2
      = 97 .33 J

97.3 J (nearest tenth)