Answer :
We start with the formula for the final velocity when falling from rest:
[tex]$$
v = \sqrt{2gh}
$$[/tex]
where
[tex]\( v = 12 \)[/tex] feet per second (speed when the hammer hits the floor)
[tex]\( g = 32 \)[/tex] feet per second squared (acceleration due to gravity)
[tex]\( h \)[/tex] is the height above the ground.
Step 1: Square both sides of the equation.
[tex]$$
v^2 = 2gh
$$[/tex]
Step 2: Solve for [tex]\( h \)[/tex].
[tex]$$
h = \frac{v^2}{2g}
$$[/tex]
Step 3: Substitute the given values.
[tex]$$
h = \frac{12^2}{2 \times 32}
$$[/tex]
Step 4: Compute [tex]\( 12^2 \)[/tex] and the denominator.
[tex]$$
12^2 = 144, \quad 2 \times 32 = 64
$$[/tex]
Step 5: Divide to find [tex]\( h \)[/tex].
[tex]$$
h = \frac{144}{64} = 2.25 \text{ feet}
$$[/tex]
Thus, the hammer was dropped from a height of [tex]$2.25$[/tex] feet above the ground.
The correct answer is therefore option D: [tex]$2.25$[/tex] feet.
[tex]$$
v = \sqrt{2gh}
$$[/tex]
where
[tex]\( v = 12 \)[/tex] feet per second (speed when the hammer hits the floor)
[tex]\( g = 32 \)[/tex] feet per second squared (acceleration due to gravity)
[tex]\( h \)[/tex] is the height above the ground.
Step 1: Square both sides of the equation.
[tex]$$
v^2 = 2gh
$$[/tex]
Step 2: Solve for [tex]\( h \)[/tex].
[tex]$$
h = \frac{v^2}{2g}
$$[/tex]
Step 3: Substitute the given values.
[tex]$$
h = \frac{12^2}{2 \times 32}
$$[/tex]
Step 4: Compute [tex]\( 12^2 \)[/tex] and the denominator.
[tex]$$
12^2 = 144, \quad 2 \times 32 = 64
$$[/tex]
Step 5: Divide to find [tex]\( h \)[/tex].
[tex]$$
h = \frac{144}{64} = 2.25 \text{ feet}
$$[/tex]
Thus, the hammer was dropped from a height of [tex]$2.25$[/tex] feet above the ground.
The correct answer is therefore option D: [tex]$2.25$[/tex] feet.
