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

To determine how far above the ground the hammer was when it was dropped, you can 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 $[/tex] is the acceleration due to gravity, and
- [tex]$ h $[/tex] is the initial height from which the object was dropped.

We are given:
- [tex]$ v = 8 $[/tex] feet per second,
- [tex]$ g = 32 $[/tex] feet per second squared.

We need to find [tex]$ h $[/tex], the height above the ground from which the hammer was dropped.

First, rearrange the formula to solve for [tex]$ h $[/tex]:

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

Square both sides to eliminate the square root:

[tex]\[ v^2 = 2gh \][/tex]

Now, solve for [tex]$ h $[/tex]:

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

Plug the given values into the equation:

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

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

[tex]\[ h = 1 \][/tex]

hence, the hammer was dropped from a height of 1.0 foot above the ground.

The correct answer is:

C. 1.0 foot