#### Answer

(a) v = 42 m/s
(b) $P = 3.2 \times 10^5 ~W$

#### Work Step by Step

(a) We can use conservation of energy to solve this question. The kinetic energy at the bottom is equal to the potential energy at the top:
$KE = PE$
$\frac{1}{2}mv^2 = mgy$
$v^2 = 2gy$
$v = \sqrt{2gy} = \sqrt{(2)(9.80 ~m/s^2)(88 ~m)}$
$v = 42 ~m/s$
(b) 55% of the initial potential energy will be transferred to the turbine blades. Note that 680 kg of water fall each second. As a result,
$P = \frac{0.55PE}{t}$
$P = \frac{0.55mgy}{t}$
$P = \frac{(0.55)(680 ~kg)(9.80 ~m/s^2)(88 ~m)}{1 ~s}$
$P = 3.2 \times 10^5 ~W$