# From approximately what floor of a building

From approximately what floor of a building must a car be dropped from an at-rest position so that it reaches a speed of 80.7 ft/s (55 mi/h) when it hits the ground? Each floor is 12 ft higher than the one below it.

#### Solution:

Let us use the following equation to find the solution to this problem:

$v_2^2=v_1^2+2a_c(s_2-s_1)$

(Where $v_2$ is final velocity, $v_1$ is initial velocity, $a_c$ is constant acceleration, $s_2$ is initial displacement, and $s_1$ is final displacement)

From the question, we have the following given:

$v_2=80.7$ ft/s

$v_1=0$ ft/s

$a_c=32.2$ ft/$s^2$

$s_1=0$ ft

Substitute these values into our equation to find $s_2$.

$v_2^2=v_1^2+2a_c(s_2-s_1)$

$80.7^2=0+2(32.3)(s_2-0)$

$s_2=101.1$ ft

The question tells us that each floor is 12 ft apart, thus to figure out the floor number, we will divide the value of $s_2$ (height distance) by 12.

floor # = $\dfrac{101.1}{12}=8.425$

As we cannot have 0.425th of a floor, we will round to the upper integer which is the 9th floor.