A 1 kg object is dropped from rest from a height of 20 m. What is its speed just before hitting the ground? (Take g = 10 m/s^2 and ignore air resistance.)
Show the answer and explanation
Correct answer
B. 20 m/s
Principle or equation
Conservation of mechanical energy: mgh = (1/2) m v^2, so v = sqrt(2gh).
Why this answer is correct
v = sqrt(2 * 10 * 20) = sqrt(400) = 20 m/s.
Example
For a height of 5 m, v = sqrt(2*10*5) = 10 m/s.
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