Air resistance determines fall time for differently sized and massed balls

Steps:

• With air resistance, balls reach terminal velocity $v_t \propto \sqrt{m / D^2}$, where $m$ is mass and $D$ is diameter (cross-sectional area $\propto D^2$).
• Normalize to option A: $v_{tA} = \sqrt{M / D^2}$.
• Option B: $v_{tB} = \sqrt{4M / D^2} = 2 v_{tA}$.
• Options A, C, D have lower $v_t$ (1, 0.5, and 1 times $v_{tA}$), so B falls fastest.

Why B is correct:

• B has the highest mass-to-area ratio, yielding maximum $v_t$ per the drag equation $v_t = \sqrt{2mg / (C_d \rho A)}$.

Why the others are wrong:

• A: Baseline; lower mass than B increases drag effect relative to inertia.
• C: Larger diameter quadruples area, halving $v_t$ despite same mass.
• D: Mass increase offset by quadrupled area, yielding same $v_t$ as A.

Final answer: B