#### Answer

$v = 1.9~m/s$

#### Work Step by Step

The work done on the fish by the tension will be equal to the sum of the kinetic energy and the potential energy. Therefore;
$KE+PE = W_T$
$\frac{1}{2}mv^2+mgh = T~d$
$v^2 = \frac{2(T~d-mgh)}{m}$
$v = \sqrt{\frac{2~(T~d-mgh)}{m}}$
$v = \sqrt{\frac{(2)~[(16~N)(2.0~m)-(1.5~kg)(9.80~m/s^2)(2.0~m)]}{1.5~kg}}$
$v = 1.9~m/s$