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:

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

(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_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.

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.

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

#### Final Answer: