How do you determine the height for Part B of this question?

Let height of the building be `=h " "m`
After the wrench is dropped from the roof top, the kinematic equation is given by
`v^2-u^2=2gs` ......(1)
where `v, u and g` are final velocity after dropping distance `s`, initial velocity and acceleration due to gravity respectively. Let `g=9.8ms^-2`

Let us find out the distance dropped when workman standing at the eighth floor observes the wrench. (for simplicity eyes of worker assumed at the floor level)
`(33.1)^2-0^2=2xx9.8xxs`
`=>s=(33.1)^2/(2xx9.8)`
`=>sapprox55.9 m`
Since the floors above the first are of height `8m` each
Hence number of floors above the eighth floor`=55.9/8`
`=7`, rounded to nearest digit as number of floors can not be a fraction.
Total number of floors of the building `=7+7=14`
We need to remember that the workman standing on the eighth floor has only seven floors below him.

Height of the building `h=12.0+13xx8.00=116.00m`
(First floor is of `12.0 m` and all other floors are of `8.00 m` height)
From (1) velocity `v` when the wrench hits the ground
`v^2=2xx9.8xx118.00`
`v=47.68ms^-1`, rounded to two decimal places.

Kinetic energy of wrench when it hits the ground`=mgh`
`=0.125xx9.8xx116=142J`

(All its potential energy while at the roof gets converted into its kinetic energy as it hits the ground.
It can also be calculated using the expression `KE=1/2mv^2`)