A 1.5 kg object is dropped from rest from a height of 12 m. What is its speed just before hitting the ground? (Ignore air resistance and take g = 10 m/s².)
Show the answer and explanation
Correct answer
B. About 15 m/s
Principle or equation
Conservation of mechanical energy: mgh = (1/2)mv², so v = sqrt(2gh).
Why this answer is correct
Set initial potential energy equal to final kinetic energy: mgh = (1/2)mv². Mass cancels: v = sqrt(2gh) = sqrt(2 × 10 × 12) = sqrt(240) ≈ 15.5 m/s, which rounds to about 15 m/s.
Example
For h = 5 m, v = sqrt(2 × 10 × 5) = sqrt(100) = 10 m/s.
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