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Answer :
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
[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
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