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. 1.0 foot
C. 2.25 feet
D. 8.5 feet

We are given that when the hammer hits the floor, its speed is [tex]$v = 12$[/tex] feet per second, and the acceleration due to gravity is [tex]$g = 32$[/tex] feet per second squared. We use the formula

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

where [tex]$h$[/tex] is the height from which the hammer was dropped.

Step 1. Square both sides of the equation

Starting with
[tex]$$
v = \sqrt{2 g h},
$$[/tex]
square both sides to obtain
[tex]$$
v^2 = 2gh.
$$[/tex]

Step 2. Solve for [tex]$h$[/tex]

Rearrange the equation to isolate [tex]$h$[/tex]:
[tex]$$
h = \frac{v^2}{2g}.
$$[/tex]

Step 3. Substitute the given values

Substitute [tex]$v = 12$[/tex] and [tex]$g = 32$[/tex] into the formula:
[tex]$$
h = \frac{12^2}{2 \times 32}.
$$[/tex]

Step 4. Compute the numerical values

First, compute [tex]$12^2$[/tex]:
[tex]$$
12^2 = 144.
$$[/tex]

Then, compute [tex]$2 \times 32$[/tex]:
[tex]$$
2 \times 32 = 64.
$$[/tex]

Now, substitute these into the equation for [tex]$h$[/tex]:
[tex]$$
h = \frac{144}{64}.
$$[/tex]

Step 5. Simplify the fraction

Divide [tex]$144$[/tex] by [tex]$64$[/tex]:
[tex]$$
h = 2.25 \text{ feet.}
$$[/tex]

Thus, the hammer was dropped from a height of [tex]$\boxed{2.25 \text{ feet}}$[/tex], which corresponds to option C.