Answer
$K = 60.3~J$
Work Step by Step
We can find the speed halfway to the ground:
$v_f^2 = v_0^2+2ay$
$v_f = \sqrt{v_0^2+2ay}$
$v_f = \sqrt{(3.00~m/s)^2+(2)(9.8~m/s^2)(2.00~m)}$
$v_f = 6.943~m/s$
We can find the kinetic energy:
$K = \frac{1}{2}mv^2$
$K = \frac{1}{2}(2.50~kg)(6.943~m/s)^2$
$K = 60.3~J$
