Answer
The sphere travels through a distance of $~~5.8~m~~$ during the deceleration period.
Work Step by Step
We can find the speed at the bottom of the tower:
$v^2 = v_0^2+2ay$
$v^2 = 0+2ay$
$v = \sqrt{2ay}$
$v = \sqrt{(2)(9.8~m/s^2)(145~m)}$
$v = 53.3~m/s$
For the next part of the question, let $v_0 = 53.3~m/s$ and let $v = 0$
We can find the distance the sphere travels during the deceleration period:
$v^2 = v_0^2+2ay$
$y = \frac{v^2 - v_0^2}{2a}$
$y = \frac{0 -(53.3~m/s)^2}{(2)(25)(-9.8~m/s^2)}$
$y = 5.8~m$
The sphere travels through a distance of $~~5.8~m~~$ during the deceleration period.
