Answer :
We start with the formula that relates the final speed [tex]$v$[/tex], the acceleration due to gravity [tex]$g$[/tex], and the height [tex]$h$[/tex] from which the hammer is dropped:
[tex]$$
v = \sqrt{2gh}.
$$[/tex]
Our goal is to find the height [tex]$h$[/tex]. We begin by squaring both sides of the equation to eliminate the square root:
[tex]$$
v^2 = 2gh.
$$[/tex]
Next, we solve for [tex]$h$[/tex] by dividing both sides by [tex]$2g$[/tex]:
[tex]$$
h = \frac{v^2}{2g}.
$$[/tex]
Now, we substitute the given values [tex]$v = 12$[/tex] feet per second and [tex]$g = 32$[/tex] feet/second[tex]$^2$[/tex] into the equation:
[tex]$$
h = \frac{12^2}{2 \times 32}.
$$[/tex]
Calculating the numerator:
[tex]$$
12^2 = 144.
$$[/tex]
Calculating the denominator:
[tex]$$
2 \times 32 = 64.
$$[/tex]
Thus, the height is:
[tex]$$
h = \frac{144}{64} = 2.25 \text{ feet}.
$$[/tex]
Therefore, the hammer was dropped from a height of [tex]$2.25$[/tex] feet, which corresponds to option C.
[tex]$$
v = \sqrt{2gh}.
$$[/tex]
Our goal is to find the height [tex]$h$[/tex]. We begin by squaring both sides of the equation to eliminate the square root:
[tex]$$
v^2 = 2gh.
$$[/tex]
Next, we solve for [tex]$h$[/tex] by dividing both sides by [tex]$2g$[/tex]:
[tex]$$
h = \frac{v^2}{2g}.
$$[/tex]
Now, we substitute the given values [tex]$v = 12$[/tex] feet per second and [tex]$g = 32$[/tex] feet/second[tex]$^2$[/tex] into the equation:
[tex]$$
h = \frac{12^2}{2 \times 32}.
$$[/tex]
Calculating the numerator:
[tex]$$
12^2 = 144.
$$[/tex]
Calculating the denominator:
[tex]$$
2 \times 32 = 64.
$$[/tex]
Thus, the height is:
[tex]$$
h = \frac{144}{64} = 2.25 \text{ feet}.
$$[/tex]
Therefore, the hammer was dropped from a height of [tex]$2.25$[/tex] feet, which corresponds to option C.
