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

To solve the problem of finding out how far above the ground the hammer was dropped from, we can use the formula:

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

where:
- [tex]$ v $[/tex] is the velocity at which the hammer hits the ground (4 feet per second),
- [tex]$ g $[/tex] is the acceleration due to gravity (32 feet/second²),
- [tex]$ h $[/tex] is the height from which the hammer was dropped.

We need to rearrange this formula 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 of the equation to eliminate 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. Substitute the known values into the equation:
- [tex]$ v = 4 $[/tex] feet/second
- [tex]$ g = 32 $[/tex] feet/second²

[tex]\[ h = \frac{4^2}{2 \times 32} \][/tex]

5. Calculate the value:
- [tex]$ 4^2 = 16 $[/tex]
- [tex]$ 2 \times 32 = 64 $[/tex]
- [tex]$ \frac{16}{64} = 0.25 $[/tex]

The height ([tex]$ h $[/tex]) from which the hammer was dropped is 0.25 feet above the ground.

Therefore, the correct answer is C. 0.25 feet.