Free fall under constant acceleration

Steps:

• The feather falls from rest with constant acceleration g in vacuum, so distance s = (1/2)gt².
• For total distance L in time T: L = (1/2)gT².
• In time 0.5T: s = (1/2)g(0.5T)² = (1/2)g(0.25T²) = 0.25 × [(1/2)gT²] = 0.25L.
• Thus, the distance fallen is 0.25L.

Why B is correct:

• The kinematic equation s = (1/2)at² shows distance scales with t², so half time yields (1/2)² = 1/4 of total distance.

Why the others are wrong:

• A: 0.1L underestimates; ignores quadratic time dependence.
• C: 0.3L incorrect; confuses with linear velocity-time relation.
• D: 0.5L wrong; assumes constant velocity, not acceleration from rest.

Final answer: B