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

Jul 7, 2016

$14$
$47.68 m {s}^{-} 1$, rounded to two decimal places.
$142 J$

#### Explanation:

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

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)
${\left(33.1\right)}^{2} - {0}^{2} = 2 \times 9.8 \times s$
$\implies s = {\left(33.1\right)}^{2} / \left(2 \times 9.8\right)$
$\implies s \approx 55.9 m$
Since the floors above the first are of height $8 m$ each
Hence number of floors above the eighth floor$= \frac{55.9}{8}$
$= 7$, 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 $h = 12.0 + 13 \times 8.00 = 116.00 m$
(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
${v}^{2} = 2 \times 9.8 \times 118.00$
$v = 47.68 m {s}^{-} 1$, rounded to two decimal places.

Kinetic energy of wrench when it hits the ground$= m g h$
$= 0.125 \times 9.8 \times 116 = 142 J$

(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 $K E = \frac{1}{2} m {v}^{2}$)