You are helping with some repairs at home. You drop a hammer and it hits the floor at a speed of 8 feet per second. If the acceleration due to gravity ([tex]$g$[/tex]) is 32 feet/second[tex]$^2$[/tex], how far above the ground ([tex]$h$[/tex]) was the hammer when you dropped it? Use the formula:

[tex]\[ v = \sqrt{2gh} \][/tex]

A. 8.0 feet
B. 16.0 feet
C. 1.0 foot
D. 2.0 feet

To find the height ( [tex]$ h $[/tex] ) from which the hammer was dropped, we can use the formula:

[tex]\[ v = \sqrt{2gh} \][/tex]

where:
- [tex]$ v $[/tex] is the final speed of the hammer when it hits the ground, which is 8 feet per second.
- [tex]$ g $[/tex] is the acceleration due to gravity, which is 32 feet per second squared.
- [tex]$ h $[/tex] is the height from which the hammer was dropped.

First, let's rearrange the formula to solve for [tex]$ h $[/tex]:

1. Square both sides to eliminate the square root:
[tex]\[
v^2 = 2gh
\][/tex]

2. Solve for [tex]$ h $[/tex]:
[tex]\[
h = \frac{v^2}{2g}
\][/tex]

Substitute the known values into the equation:

- [tex]$ v = 8 $[/tex] feet per second
- [tex]$ g = 32 $[/tex] feet per second squared

3. Plug in these values:
[tex]\[
h = \frac{8^2}{2 \times 32}
\][/tex]

4. Calculate [tex]$ 8^2 $[/tex]:
[tex]\[
8^2 = 64
\][/tex]

5. Calculate [tex]$ 2 \times 32 $[/tex]:
[tex]\[
2 \times 32 = 64
\][/tex]

6. Now, divide the results:
[tex]\[
h = \frac{64}{64} = 1.0
\][/tex]

Therefore, the hammer was dropped from a height of 1.0 foot. The correct answer is C. 1.0 foot.