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 solve this problem, we need to find out how high above the ground the hammer was when it was dropped. We can use the formula for the speed of an object falling under gravity, which is given as:

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

where:
- [tex]\( v \)[/tex] is the speed of the object when it hits the ground (8 feet per second in this case),
- [tex]\( g \)[/tex] is the acceleration due to gravity (32 feet/second² here),
- [tex]\( h \)[/tex] is the height from which the object was dropped.

We are asked to find [tex]\( h \)[/tex]. We begin by rearranging the formula to solve for [tex]\( h \)[/tex]:

1. Square both sides of the equation to eliminate 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]

Now, plug in the given values:

- [tex]\( v = 8 \)[/tex] feet/second
- [tex]\( g = 32 \)[/tex] feet/second²

[tex]\[ h = \frac{8^2}{2 \cdot 32} \][/tex]

[tex]\[ h = \frac{64}{64} \][/tex]

[tex]\[ h = 1.0 \][/tex]

So, the hammer was 1.0 foot above the ground when it was dropped. Therefore, the correct answer is option A: 1.0 foot.