Answer
The speed at the bottom of the tower is $~~53.3~m/s$
Work Step by Step
We can find the speed at the bottom of the tower:
$v^2 = v_0^2+2ay$
$v^2 = 0+2ay$
$v = \sqrt{2ay}$
$v = \sqrt{(2)(9.8~m/s^2)(145~m)}$
$v = 53.3~m/s$
The speed at the bottom of the tower is $~~53.3~m/s$.
