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------------------------------------------------ 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. 8.0 feet
C. 2.0 feet
D. 16.0 feet

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

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

Where:
- [tex]\( v \)[/tex] is the speed of the hammer when it hits the floor, 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 from which the hammer was dropped.

First, we need to solve for [tex]\( h \)[/tex]. We can rearrange the formula to:

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

Now, solve for [tex]\( h \)[/tex] by dividing both sides by [tex]\( 2g \)[/tex]:

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

Substitute the known values into the equation:

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

Calculate the square of the speed:

[tex]\[ 8^2 = 64 \][/tex]

Now substitute that into the equation:

[tex]\[ h = \frac{64}{2 \times 32} \][/tex]

Calculate the denominator:

[tex]\[ 2 \times 32 = 64 \][/tex]

Now divide the numerator by the denominator:

[tex]\[ h = \frac{64}{64} = 1 \][/tex]

So, the hammer was dropped from a height of 1.0 foot.

The correct answer is:

A. 1.0 foot