A ball of mass 1.5 kg is dropped from a height of 12 m…

Step 1. Total. $1.5\ \text{kg} \times 9.8\ \text{m/s}^2 \times 12\ \text{m} = 176.4\ \text{J}$.

Step 2. (a) At $6\ \text{m}$: $PE = 1.5\ \text{kg} \times 9.8\ \text{m/s}^2 \times 6\ \text{m} = 88.2\ \text{J}$.

Step 3. (b) $KE = 176.4\ \text{J} - 88.2\ \text{J} = 88.2\ \text{J}$.

Step 4. (c) It has fallen $6\ \text{m}$: $v = \sqrt{2 \times 9.8\ \text{m/s}^2 \times 6\ \text{m}} = 10.84\ \text{m/s}$.