Answer
The correct answer is (b) $0.012~m/s$.
Work Step by Step
The momentum of the fish $p_f$ will be equal in magnitude to the momentum of the water drop $p_w$.
$p_f = p_w$
$m_f~v_f = m_w~v_w$
$v_f = \frac{m_w~v_w}{m_f}$
$v_f = \frac{(3\times 10^{-4}~kg)(2.5~m/s)}{0.065~kg}$
$v_f = 0.012~m/s$
The correct answer is (b) $0.012~m/s$.
