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Answer :
We start with the formula
[tex]$$
v = \sqrt{2gh}.
$$[/tex]
Squaring both sides gives
[tex]$$
v^2 = 2gh.
$$[/tex]
Next, solving for [tex]$h$[/tex], we have
[tex]$$
h = \frac{v^2}{2g}.
$$[/tex]
Substitute the given values [tex]$v = 8$[/tex] feet per second and [tex]$g = 32$[/tex] feet per second squared:
[tex]$$
h = \frac{8^2}{2 \times 32} = \frac{64}{64} = 1.0 \text{ foot}.
$$[/tex]
Thus, the hammer was dropped from a height of
[tex]$$
\boxed{1.0 \text{ foot}}.
$$[/tex]
The correct answer is Option A.
[tex]$$
v = \sqrt{2gh}.
$$[/tex]
Squaring both sides gives
[tex]$$
v^2 = 2gh.
$$[/tex]
Next, solving for [tex]$h$[/tex], we have
[tex]$$
h = \frac{v^2}{2g}.
$$[/tex]
Substitute the given values [tex]$v = 8$[/tex] feet per second and [tex]$g = 32$[/tex] feet per second squared:
[tex]$$
h = \frac{8^2}{2 \times 32} = \frac{64}{64} = 1.0 \text{ foot}.
$$[/tex]
Thus, the hammer was dropped from a height of
[tex]$$
\boxed{1.0 \text{ foot}}.
$$[/tex]
The correct answer is Option A.
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