Discussion
A man jumps from a height in a deep pool of water. If the net frictional force of water F is proportional to the instantaneous speed V of the man, i.e., F = kV; what is the terminal velocity of the man ? The mass of the man is m.
Answer: A
The terminal velocity is \( mg/k \).
The man accelerates until the resisting force balances his weight; then the net force is zero and the speed stays constant:
\[ kV_{t} = mg \quad \Rightarrow \quad V_{t} = \frac{mg}{k} \]
Solving \( m\,dV/dt = mg - kV \) gives \( V = \frac{mg}{k}(1 - e^{-kt/m}) \), which approaches the same value. (Buoyancy is ignored, as the question does.)
The multiples \( 2mg/k \), \( \frac{1}{2}mg/k \) and \( \frac{3}{2}mg/k \) do not satisfy the force balance for a resistance that is simply \( kV \).
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