Answer
The speed at the bottom of the cliff is 173 m/s.
Work Step by Step
The kinetic energy at the bottom will be equal to the kinetic energy plus the potential energy at the top. Therefore,
$KE_2 = KE_1 + PE$
$\frac{1}{2}mv_2^2 = \frac{1}{2}mv_1^2 + mgh$
$v_2 = \sqrt{v_1^2 + 2gh}$
$v_2 = \sqrt{(165 ~m/s)^2 + (2)(9.8 ~m/s^2)(135 ~m)}$
$v_2 = 173 ~m/s$
The speed at the bottom of the cliff is 173 m/s.