1. **Stating the problem:** A volleyball with mass 240 g is thrown downward from a height of 10 m with an initial velocity of 2.5 m/s. Given gravitational acceleration $g = 10$ m/s$^2$, find the kinetic energy at a height of 3 m. 2. **Formula used:** The total mechanical energy at any height is the sum of kinetic energy (KE) and potential energy (PE): $$E = KE + PE$$ Potential energy at height $h$ is: $$PE = mgh$$ Kinetic energy is: $$KE = \frac{1}{2}mv^2$$ Since mechanical energy is conserved (ignoring air resistance), total energy at 10 m equals total energy at 3 m: $$KE_{10} + PE_{10} = KE_3 + PE_3$$ 3. **Calculate initial kinetic energy at 10 m:** Mass $m = 240$ g = 0.24 kg Initial velocity $v_0 = 2.5$ m/s $$KE_{10} = \frac{1}{2} \times 0.24 \times (2.5)^2 = 0.12 \times 6.25 = 0.75 \text{ J}$$ 4. **Calculate potential energy at 10 m:** $$PE_{10} = 0.24 \times 10 \times 10 = 24 \text{ J}$$ 5. **Total mechanical energy at 10 m:** $$E = KE_{10} + PE_{10} = 0.75 + 24 = 24.75 \text{ J}$$ 6. **Calculate potential energy at 3 m:** $$PE_3 = 0.24 \times 10 \times 3 = 7.2 \text{ J}$$ 7. **Find kinetic energy at 3 m:** $$KE_3 = E - PE_3 = 24.75 - 7.2 = 17.55 \text{ J}$$ **Final answer:** The kinetic energy of the volleyball at 3 m height is **17.55 J**.