一个质量为1 kg的小球从离地20 m高处自由下落，不计空气阻力，重力加速度取10 m/s²。求小球落地时的速度大小。
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Correct answer
C. 20 m/s
Principle or equation
机械能守恒：mgh = 1/2 mv²，所以 v = sqrt(2gh)。
Why this answer is correct
v = sqrt(2 * 10 * 20) = sqrt(400) = 20 m/s。
Example
若高度为5 m，则 v = sqrt(100) = 10 m/s。
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