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 [tex]$/$[/tex] second [tex]${ }^2$[/tex], how far above the ground [tex]$(h)$[/tex] was the hammer when you dropped it? Use the formula:

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

A. 8.5 feet
B. 1.0 foot
C. 18.0 feet
D. 2.25 feet

Sure! Let's solve the problem step by step:

We are given the following information:

- The final velocity of the hammer when it hits the ground, [tex]$ v = 12 $[/tex] feet per second.
- The acceleration due to gravity, [tex]$ g = 32 $[/tex] feet per second squared.

We need to find the height [tex]$ h $[/tex] from which the hammer was dropped. We can use the formula:
[tex]\[ v = \sqrt{2gh} \][/tex]

First, let's rearrange the formula to solve for [tex]$ h $[/tex]:
[tex]\[ v^2 = 2gh \][/tex]

Next, we divide both sides by [tex]$ 2g $[/tex]:
[tex]\[ h = \frac{v^2}{2g} \][/tex]

Now we can substitute the given values into the formula:
- [tex]$ v = 12 $[/tex] feet per second
- [tex]$ g = 32 $[/tex] feet per second squared

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

Calculate the value step by step:

1. [tex]$ 12^2 = 144 $[/tex]
2. [tex]$ 2 \times 32 = 64 $[/tex]
3. [tex]$ \frac{144}{64} = 2.25 $[/tex]

Therefore, the height [tex]$ h $[/tex] from which the hammer was dropped is:
[tex]\[ h = 2.25 \text{ feet} \][/tex]

Thus, the correct answer is:
D. 2.25 feet