You are helping with some repairs at home. You drop a hammer, and it hits the floor at a speed of 12 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. 18.0 feet
B. 8.5 feet
C. 1.0 foot
D. 2.25 feet

To find the height from which the hammer was dropped, we can use the formula for the relationship between speed, gravity, and height:

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

where:
- [tex]$ v $[/tex] is the final speed of the hammer (12 feet per second),
- [tex]$ g $[/tex] is the acceleration due to gravity (32 feet/second[tex]$^2$[/tex]),
- [tex]$ h $[/tex] is the height from which the hammer was dropped.

We need to solve for [tex]$ h $[/tex]. Here's how you can do it step-by-step:

1. Start with the formula:

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

2. Square both sides to get rid of the square root:

[tex]\[ v^2 = 2gh \][/tex]

3. Solve for [tex]$ h $[/tex] by dividing both sides by [tex]$ 2g $[/tex]:

[tex]\[ h = \frac{v^2}{2g} \][/tex]

4. Plug in the known values ([tex]$ v = 12 $[/tex] feet/second and [tex]$ g = 32 $[/tex] feet/second[tex]$^2$[/tex]):

[tex]\[ h = \frac{12^2}{2 \times 32} \][/tex]

5. Calculate:

[tex]\[ h = \frac{144}{64} \][/tex]
[tex]\[ h = 2.25 \][/tex]

So, the hammer was dropped from a height of 2.25 feet above the ground. Thus, the correct answer is:

D. 2.25 feet