You are helping with some repairs at home. You drop a hammer and it hits the floor at a speed of 4 feet per second. If the acceleration due to gravity ([tex]g[/tex]) is 32 feet per second squared, how far above the ground ([tex]h[/tex]) was the hammer when you dropped it? Use the formula:

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

A. 16.0 feet
B. 0.5 feet
C. 1.0 foot
D. 0.25 feet

Sure, let's solve this problem step by step using the given formula:

Given:
- The speed at which the hammer hits the floor, [tex]$ v = 4 $[/tex] feet per second.
- The acceleration due to gravity, [tex]$ g = 32 $[/tex] feet/second[tex]$^2$[/tex].
- The formula to find the height from which the hammer was dropped: [tex]$ v = \sqrt{2gh} $[/tex].

We need to find the height [tex]$ h $[/tex]. Here is the step-by-step process:

1. Write down the given formula:
[tex]\[
v = \sqrt{2 g h}
\][/tex]

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

3. Plug the given values of [tex]$ v $[/tex] and [tex]$ g $[/tex] into the equation:
[tex]\[
4^2 = 2 \cdot 32 \cdot h
\][/tex]

4. Simplify the equation:
[tex]\[
16 = 64h
\][/tex]

5. Solve for [tex]$ h $[/tex] by dividing both sides of the equation by 64:
[tex]\[
h = \frac{16}{64}
\][/tex]

6. Simplify the fraction:
[tex]\[
h = 0.25 \text{ feet}
\][/tex]

Thus, the height from which the hammer was dropped is [tex]$\boxed{0.25 \text{ feet}}$[/tex].

So, the correct answer is:
D. 0.25 feet