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Answer :
To find out how far above the ground the hammer was when it was dropped, we can use the formula for velocity when an object is dropped from a certain height:
[tex]\[ v = \sqrt{2gh} \][/tex]
where [tex]\( v \)[/tex] is the speed at which the hammer hits the floor, [tex]\( g \)[/tex] is the acceleration due to gravity, and [tex]\( h \)[/tex] is the height from which the hammer was dropped.
We have:
- [tex]\( v = 8 \)[/tex] feet per second
- [tex]\( g = 32 \)[/tex] feet per second squared
We need to solve for [tex]\( h \)[/tex].
First, square both sides of the equation to eliminate the square root:
[tex]\[ v^2 = 2gh \][/tex]
Substitute the known values:
[tex]\[ 8^2 = 2 \times 32 \times h \][/tex]
Calculate [tex]\( 8^2 \)[/tex]:
[tex]\[ 64 = 2 \times 32 \times h \][/tex]
Simplify the right side:
[tex]\[ 64 = 64h \][/tex]
Divide both sides by 64 to solve for [tex]\( h \)[/tex]:
[tex]\[ h = 1 \][/tex]
So, the hammer was dropped from a height of 1.0 feet.
The correct answer is 1.0 feet.
[tex]\[ v = \sqrt{2gh} \][/tex]
where [tex]\( v \)[/tex] is the speed at which the hammer hits the floor, [tex]\( g \)[/tex] is the acceleration due to gravity, and [tex]\( h \)[/tex] is the height from which the hammer was dropped.
We have:
- [tex]\( v = 8 \)[/tex] feet per second
- [tex]\( g = 32 \)[/tex] feet per second squared
We need to solve for [tex]\( h \)[/tex].
First, square both sides of the equation to eliminate the square root:
[tex]\[ v^2 = 2gh \][/tex]
Substitute the known values:
[tex]\[ 8^2 = 2 \times 32 \times h \][/tex]
Calculate [tex]\( 8^2 \)[/tex]:
[tex]\[ 64 = 2 \times 32 \times h \][/tex]
Simplify the right side:
[tex]\[ 64 = 64h \][/tex]
Divide both sides by 64 to solve for [tex]\( h \)[/tex]:
[tex]\[ h = 1 \][/tex]
So, the hammer was dropped from a height of 1.0 feet.
The correct answer is 1.0 feet.
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