You are helping with some repairs at home. You drop a hammer, and it hits the floor at a speed of 4 feet per second. If the acceleration due to gravity [tex]$ (g) $[/tex] is 32 feet/second[tex]$^2$[/tex], how far above the ground [tex]$(h)$[/tex] was the hammer when you dropped it? Use the formula:

[tex]\[ v = \sqrt{2gh} \][/tex]

A. 1.0 foot
B. 0.25 feet
C. 0.5 feet
D. 16.0 feet

We are given the formula

[tex]$$
v = \sqrt{2gh}
$$[/tex]

where
[tex]$ v = 4 $[/tex] ft/s (the speed when it hits the ground) and
[tex]$ g = 32 $[/tex] ft/s[tex]$^2$[/tex] (the acceleration due to gravity).

Step 1: Square both sides

Square both sides of the equation to eliminate the square root:

[tex]$$
v^2 = 2gh.
$$[/tex]

Step 2: Solve for [tex]$ h $[/tex]

Isolate [tex]$ h $[/tex] by dividing both sides by [tex]$ 2g $[/tex]:

[tex]$$
h = \frac{v^2}{2g}.
$$[/tex]

Step 3: Substitute the given values

Substitute [tex]$ v = 4 $[/tex] ft/s and [tex]$ g = 32 $[/tex] ft/s[tex]$^2$[/tex] into the expression:

[tex]$$
h = \frac{4^2}{2 \times 32}.
$$[/tex]

First compute [tex]$ 4^2 $[/tex]:

[tex]$$
4^2 = 16.
$$[/tex]

Then compute the denominator:

[tex]$$
2 \times 32 = 64.
$$[/tex]

So,

[tex]$$
h = \frac{16}{64}.
$$[/tex]

Step 4: Simplify the fraction

Simplify [tex]$ \frac{16}{64} $[/tex]:

[tex]$$
h = 0.25 \text{ feet}.
$$[/tex]

Thus, the hammer was dropped from a height of [tex]$ 0.25 $[/tex] feet above the ground.

The correct answer is B. 0.25 feet.