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\(^2\), how far above the ground \([tex]h\)[/tex] was the hammer when you dropped it? Use the formula:

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

A. 8.0 feet
B. 16.0 feet
C. 1.0 foot
D. 2.0 feet

To solve the problem of how far above the ground the hammer was when it was dropped, we need to use the formula for the velocity of an object in free fall:

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

where:
- [tex]\( v \)[/tex] is the velocity of the object when it hits the ground,
- [tex]\( g \)[/tex] is the acceleration due to gravity,
- [tex]\( h \)[/tex] is the height from which the object was dropped.

We are given:
- [tex]\( v = 8 \)[/tex] feet per second (the speed of the hammer when it hits the floor),
- [tex]\( g = 32 \)[/tex] feet/second[tex]\(^2\)[/tex] (the acceleration due to gravity).

Our goal is to find [tex]\( h \)[/tex], the height from which the hammer was dropped. We can rearrange the formula to solve for [tex]\( h \)[/tex]:

[tex]\[ h = \frac{v^2}{2g} \][/tex]

Let's calculate the height [tex]\( h \)[/tex]:

1. Square the velocity: [tex]\( v^2 = 8^2 = 64 \)[/tex].
2. Multiply the gravity by 2: [tex]\( 2g = 2 \times 32 = 64 \)[/tex].
3. Divide the squared velocity by twice the gravity:

[tex]\[ h = \frac{64}{64} = 1.0 \][/tex]

Therefore, the height from which the hammer was dropped is 1.0 foot. The correct answer is:

C. 1.0 foot