Explanation:
At the top of the the ball's trajectory, there is only the horizontal component of the initial velocity, which is [tex]v_0\cos30,[/tex] so the kinetic energy of the ball at this point is
[tex]KE = \frac{1}{2}m(v_0\cos30)^2[/tex]
[tex]\;\;\;\;\;= \frac{1}{2}(0.15\:\text{kg})[(40\:\text{m/s})\cos30]^2[/tex]
[tex]\;\;\;\;\;= 90\:\text{J}[/tex]
