Answer
$U = 41.0~J$
Work Step by Step
Let "down" be the negative direction.
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 height at $t = 0.448~s$:
$y = y_0+v_0~t+\frac{1}{2}at^2$
$y = (4.00~m)+(-3.00~m/s)(0.448~s)+\frac{1}{2}(-9.8~m/s^2)(0.448~s)^2$
$y = 1.673~m$
We can find the gravitational potential energy:
$U = mgh$
$U = (2.50~kg)(9.8~m/s^2)(1.673~m)$
$U = 41.0~J$
