Energy conservation links rebound height to speed Steps:

• Drop height h gives impact speed v = √(2gh).
• Rebound height is h/2, so potential energy at peak: mg(h/2) = (1/2)mu².
• Solve for u: u² = gh, so u = √(gh).
• Substitute gh = v²/2: u = v/√2.

Why B is correct:

• u/v = √(h'/h) = √(1/2) = 1/√2 from energy conservation.

Why the others are wrong:

• A: u = v/2 implies (u/v)² = 1/4, so rebound height h/4, not h/2.
• C: Same as A, incorrect height ratio.
• D: Same as A, incorrect height ratio.

Final answer: B