Terminal velocity from balancing gravitational and drag forces

Steps:

• Terminal velocity $v_t$ occurs when net force is zero: $F = 0$.
• Set $mg - k v_t^2 = 0$, so $v_t^2 = \frac{mg}{k}$.
• Solve for $v_t$: $v_t = \sqrt{\frac{mg}{k}}$.
• Thus, $v_t \propto \sqrt{m}$, a concave-down curve increasing with mass.

Why B is correct:

• B shows $v_t$ increasing as $\sqrt{m}$, matching $$ from force balance.

Why the others are wrong:

• A assumes linear $v_t \propto m$, ignoring square-root dependence on drag.
• C implies constant $v_t$, contradicting mass effect on gravity.
• D suggests inverse relation, opposite to gravity dominating for larger $m$.

Final answer: B