Answer :
To solve for the height from which the hammer was dropped, we can use the formula that relates velocity, acceleration due to gravity, and height:
[tex]\[ v = \sqrt{2gh} \][/tex]
where:
- [tex]\( v \)[/tex] is the final velocity (12 feet per second in this case),
- [tex]\( g \)[/tex] is the acceleration due to gravity (32 feet per second squared),
- [tex]\( h \)[/tex] is the height above the ground.
We need to find [tex]\( h \)[/tex]. Start by squaring both sides of the formula to eliminate the square root:
[tex]\[ v^2 = 2gh \][/tex]
Next, solve for [tex]\( h \)[/tex] by dividing both sides by [tex]\( 2g \)[/tex]:
[tex]\[ h = \frac{v^2}{2g} \][/tex]
Now, plug in the given values:
- [tex]\( v = 12 \)[/tex] feet per second,
- [tex]\( g = 32 \)[/tex] feet per second squared.
[tex]\[ h = \frac{12^2}{2 \times 32} \][/tex]
[tex]\[ h = \frac{144}{64} \][/tex]
[tex]\[ h = 2.25 \][/tex]
Therefore, the hammer was dropped from a height of 2.25 feet above the ground. The correct answer is:
D. 2.25 feet
[tex]\[ v = \sqrt{2gh} \][/tex]
where:
- [tex]\( v \)[/tex] is the final velocity (12 feet per second in this case),
- [tex]\( g \)[/tex] is the acceleration due to gravity (32 feet per second squared),
- [tex]\( h \)[/tex] is the height above the ground.
We need to find [tex]\( h \)[/tex]. Start by squaring both sides of the formula to eliminate the square root:
[tex]\[ v^2 = 2gh \][/tex]
Next, solve for [tex]\( h \)[/tex] by dividing both sides by [tex]\( 2g \)[/tex]:
[tex]\[ h = \frac{v^2}{2g} \][/tex]
Now, plug in the given values:
- [tex]\( v = 12 \)[/tex] feet per second,
- [tex]\( g = 32 \)[/tex] feet per second squared.
[tex]\[ h = \frac{12^2}{2 \times 32} \][/tex]
[tex]\[ h = \frac{144}{64} \][/tex]
[tex]\[ h = 2.25 \][/tex]
Therefore, the hammer was dropped from a height of 2.25 feet above the ground. The correct answer is:
D. 2.25 feet
