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Answer :
To solve this problem, we're going to use the formula given:
[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 above the ground from which the hammer was dropped.
First, let's rearrange the formula to solve for [tex]\( h \)[/tex]:
1. Start with the formula:
[tex]\[ v = \sqrt{2gh} \][/tex]
2. Square both sides to get rid of the square root:
[tex]\[ v^2 = 2gh \][/tex]
3. Solve for [tex]\( h \)[/tex]:
[tex]\[ h = \frac{v^2}{2g} \][/tex]
Now, plug in the given values:
- [tex]\( v = 8 \text{ feet/second} \)[/tex]
- [tex]\( g = 32 \text{ feet/second}^2 \)[/tex]
Calculate [tex]\( h \)[/tex]:
[tex]\[ h = \frac{8^2}{2 \times 32} = \frac{64}{64} = 1 \text{ foot} \][/tex]
Thus, the hammer was dropped from a height of 1 foot. So, the correct answer is:
C. 1.0 foot
[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 above the ground from which the hammer was dropped.
First, let's rearrange the formula to solve for [tex]\( h \)[/tex]:
1. Start with the formula:
[tex]\[ v = \sqrt{2gh} \][/tex]
2. Square both sides to get rid of the square root:
[tex]\[ v^2 = 2gh \][/tex]
3. Solve for [tex]\( h \)[/tex]:
[tex]\[ h = \frac{v^2}{2g} \][/tex]
Now, plug in the given values:
- [tex]\( v = 8 \text{ feet/second} \)[/tex]
- [tex]\( g = 32 \text{ feet/second}^2 \)[/tex]
Calculate [tex]\( h \)[/tex]:
[tex]\[ h = \frac{8^2}{2 \times 32} = \frac{64}{64} = 1 \text{ foot} \][/tex]
Thus, the hammer was dropped from a height of 1 foot. So, the correct answer is:
C. 1.0 foot
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