A ball is dropped from rest from height h above the ground in frame...

In frame $S$ (Earth's frame):

Initial velocity of the ball, $v_{i} = 0$

Final velocity of the ball just before hitting the ground, $v_{f} = \sqrt{2gh}$ (downwards)

Change in kinetic energy in frame $S$ is:

$\Delta K_{S} = \dfrac{1}{2} m v_{f}^2 - \dfrac{1}{2} m v_{i}^2 = \dfrac{1}{2} m (2gh) - 0 = mgh$

In frame $S'$ (moving upwards with constant speed $u$):

Initial velocity of the ball, $v'_{i} = u = v$ (downwards)

Final velocity of the ball, $v'_{f} = \sqrt{2gh} + v$ (downwards)

Change in kinetic energy in frame $S'$ is:

$\Delta K_{S'} = \dfrac{1}{2} m (v'_{f})^2 - \dfrac{1}{2} m (v'_{i})^2$

$\Delta K_{S'} = \dfrac{1}{2} m (\sqrt{2gh} + v)^2 - \dfrac{1}{2} m v^2$

$\Delta K_{S'} = \dfrac{1}{2} m (2gh + v^2 + 2v\sqrt{2gh}) - \dfrac{1}{2} m v^2$

$\Delta K_{S'} = mgh + mv\sqrt{2gh}$

Comparing the two expressions, we get $\Delta K_{S'} = \Delta K_{S} + mv\sqrt{2gh}$.

Since $m, v, g, h$ are all positive quantities, $\Delta K_{S'} > \Delta K_{S}$.

Therefore, the change in kinetic energy is larger in frame $S'$.

Answer: $\Delta K$ is larger in frame $S'$.