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{2 g h}[/tex]

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

Sure! Let's solve the problem step-by-step using the given formula:

We have the formula:

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

Where:
- [tex]$ v $[/tex] is the velocity at which the hammer hits the ground, which is 8 feet per second.
- [tex]$ g $[/tex] is the acceleration due to gravity, which is 32 feet per second squared.
- [tex]$ h $[/tex] is the height above the ground from which the hammer was dropped.

We need to find [tex]$ h $[/tex].

Firstly, let's rearrange the formula to solve for [tex]$ h $[/tex]:

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

Squaring both sides to get rid of the square root:

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

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

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

We know that:
- [tex]$ v = 8 $[/tex] feet per second
- [tex]$ g = 32 $[/tex] feet per second squared

Plug these values into the equation:

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

Calculating the values:

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

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

So, the height [tex]$ h $[/tex] from which the hammer was dropped is:

[tex]\[ \boxed{1.0 \text{ foot}} \][/tex]

Therefore, the correct answer is:
B. 1.0 foot