A body is dropped from height h. Just before it hits the ground, its speed is:
Av = g h
Bv = √(g h)
Cv = √(2 g h)
Dv = 2 g h
Answer & Solution
Correct answer: C. v = √(2 g h)
1. By energy conservation: KE_final = PE_initial.
2. ½ m v² = m g h.
3. v² = 2 g h ⇒ v = √(2 g h).
4. Same result follows from kinematics v² = u² + 2 a s with u = 0, a = g, s = h.
_Source: NCERT Class 11 Physics Ch 5, §5.7 Conservation of Mechanical Energy_
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