Answer
$K = 68.3~J$
Work Step by Step
We can find the speed at the end of the fall:
$v_f^2 = v_0^2+2ay$
$v_f = \sqrt{v_0^2+2ay}$
$v_f = \sqrt{(3.00~m/s)^2+(2)(9.8~m/s^2)(4.00~m)}$
$v_f = 9.35~m/s$
We can find the time to reach the ground:
$v_f = v_0+at$
$t = \frac{v_f-v_0}{a}$
$t = \frac{9.35~m/s-3.00~m/s}{9.8~m/s^2}$
$t = 0.648~s$
$0.200~s$ before reaching the ground, the time is $t = 0.448~s$
We can find the speed at $t = 0.448~s$:
$v_f = v_0+at$
$v_f = (3.00~m/s)+(9.8~m/s^2)(0.448~s)$
$v_f = 7.39~m/s$
We can find the kinetic energy:
$K = \frac{1}{2}mv^2$
$K = \frac{1}{2}(2.50~kg)(7.39~m/s)^2$
$K = 68.3~J$
