You can use either method, conservation of energy or kinematics, and both will give the same answer:
CONSERVATION OF ENERGY
Let's arbitrarily define downwards displacements, velocities, accelerations and forces as negative.
Falling motion:
$$mgh_1 = \frac{1}{2}mv_1^2$$
$$\therefore v_1 = -\sqrt{2gh_1}$$
Rising motion:
$$\frac{1}{2}mv_2^2 = mgh_2$$
$$\therefore v_2 = \sqrt{2gh_2}$$
$$I = m(v_2 - v_1) = m(\sqrt{2gh_2}+\sqrt{2gh_1})$$
KINEMATICS
Constant acceleration, therefore we can use suvat equations.
$$v^2 = u^2 +2as$$
Falling:
$$v_1^2 = 0 + 2(-g)(-h_1)$$
$$\therefore v_1 = -\sqrt{2gh_1}$$
Rising:
$$0 = v_2^2 + 2(-g)(h_2)$$
$$\therefore v_2 = \sqrt{2gh_2}$$
$$I = m(v_2 - v_1) = m(\sqrt{2gh_2}+\sqrt{2gh_1})$$
The reason why you get the same answers either way is because the definition of KE derived from kinematics.
The kinetic energy is the amount of work require to bring a mass, $m$ to a velocity $v$ from rest.
$$KE = work = Fs = mas$$
suvat equation:
$$v^2 = u^2 +2as$$
$$\therefore v^2 = 0 + 2as$$
Substitute:
$$v^2 = \frac{2KE}{m}$$
$$\therefore KE = \frac{1}{2}mv^2$$