You are helping with some repairs at home. You drop a hammer, and it hits the floor at a speed of 8 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. 1.0 foot
B. 2.0 feet
C. 16.0 feet
D. 8.0 feet

To find out how far above the ground the hammer was when you dropped it, you can use the formula:

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

where:
- [tex]$ v $[/tex] is the velocity at which the hammer hits the floor (8 feet per second),
- [tex]$ g $[/tex] is the acceleration due to gravity (32 feet/second²),
- [tex]$ h $[/tex] is the height above the ground that we want to find.

We'll follow these steps to solve for [tex]$ h $[/tex]:

1. Rearrange the formula to solve for [tex]$ h $[/tex]:
[tex]\[
v = \sqrt{2gh}
\][/tex]
First, square both sides to get rid of the square root:
[tex]\[
v^2 = 2gh
\][/tex]

2. Solve for [tex]$ h $[/tex] by dividing both sides by [tex]$ 2g $[/tex]:
[tex]\[
h = \frac{v^2}{2g}
\][/tex]

3. Substitute the known values into the equation:
- [tex]$ v = 8 $[/tex] feet per second,
- [tex]$ g = 32 $[/tex] feet/second².

Plug these values in:
[tex]\[
h = \frac{8^2}{2 \times 32}
\][/tex]

4. Calculate:
[tex]\[
h = \frac{64}{64}
\][/tex]
[tex]\[
h = 1.0 \text{ foot}
\][/tex]

So, the hammer was dropped from a height of 1.0 foot above the ground. The answer is A. 1.0 foot.