A 0.50 kg ball is dropped from rest from a height of 20 meters above the ground. Neglecting air resistance and taking g = 9.8 m/s^2, what is the kinetic energy of the ball just before it strikes the ground?

Answer and explanation

Correct answer: C. 98 J

  1. Step 1: By the Law of Conservation of Mechanical Energy: E_initial = E_final.
  2. Step 2: At release, the ball has zero initial kinetic energy and maximum gravitational potential energy: PE_initial = m g h.
  3. Step 3: Substitute values: PE_initial = 0.50 kg 9.8 m/s^2 20 m = 98 J.
  4. Step 4: Just before impact, all potential energy is converted to kinetic energy: KE_final = 98 J.
Answer
98 J.
Why not the others: '196 J' omits multiplying by the mass of 0.50 kg (using m = 1.0 kg); '49 J' divides by an extra factor of 2; '392 J' squares the height.

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