一物体从高为5 m的光滑斜面顶端由静止滑下，斜面倾角为30°。求物体滑到底端时的速度大小（不计空气阻力，g取10 m/s²）。
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Correct answer
B. 10 m/s
Principle or equation
机械能守恒定律：mgh = (1/2)mv²，解得 v = √(2gh)。
Why this answer is correct
v = √(2×10×5) = √100 = 10 m/s。注意速度与斜面倾角无关。
Example
若高度为20 m，则 v = √(2×10×20) = 20 m/s。
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