Answer :
We are given the formula
[tex]$$
v = \sqrt{2gh},
$$[/tex]
where:
- [tex]$v$[/tex] is the final speed (12 ft/s),
- [tex]$g$[/tex] is the acceleration due to gravity (32 ft/s[tex]$^2$[/tex]), and
- [tex]$h$[/tex] is the height.
To find [tex]$h$[/tex], we first square both sides of the equation:
[tex]$$
v^2 = 2gh.
$$[/tex]
Now, solve for [tex]$h$[/tex] by dividing both sides by [tex]$2g$[/tex]:
[tex]$$
h = \frac{v^2}{2g}.
$$[/tex]
Substitute the given values:
[tex]$$
v^2 = 12^2 = 144,
$$[/tex]
and
[tex]$$
2g = 2 \times 32 = 64.
$$[/tex]
Thus,
[tex]$$
h = \frac{144}{64} = 2.25 \text{ feet}.
$$[/tex]
So, the hammer was dropped from a height of [tex]$\boxed{2.25\ \text{feet}}$[/tex], which corresponds to option A.
[tex]$$
v = \sqrt{2gh},
$$[/tex]
where:
- [tex]$v$[/tex] is the final speed (12 ft/s),
- [tex]$g$[/tex] is the acceleration due to gravity (32 ft/s[tex]$^2$[/tex]), and
- [tex]$h$[/tex] is the height.
To find [tex]$h$[/tex], we first square both sides of the equation:
[tex]$$
v^2 = 2gh.
$$[/tex]
Now, solve for [tex]$h$[/tex] by dividing both sides by [tex]$2g$[/tex]:
[tex]$$
h = \frac{v^2}{2g}.
$$[/tex]
Substitute the given values:
[tex]$$
v^2 = 12^2 = 144,
$$[/tex]
and
[tex]$$
2g = 2 \times 32 = 64.
$$[/tex]
Thus,
[tex]$$
h = \frac{144}{64} = 2.25 \text{ feet}.
$$[/tex]
So, the hammer was dropped from a height of [tex]$\boxed{2.25\ \text{feet}}$[/tex], which corresponds to option A.
