High School

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You are helping with some repairs at home. You drop a hammer, and it hits the floor at a speed of 12 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]\[ v = \sqrt{2 g h} \][/tex]

A. 1.0 foot
B. 2.25 feet
C. 8.5 feet
D. 18.0 feet

Answer :

We start with the formula for the speed when an object falls freely:

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

Given that the final speed is [tex]$v = 12$[/tex] feet per second and the acceleration due to gravity is [tex]$g = 32$[/tex] feet per second squared, we first square the speed:

[tex]$$
v^2 = 12^2 = 144
$$[/tex]

Next, multiply the gravitational acceleration by [tex]$2$[/tex]:

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

Now, solve for the height [tex]$h$[/tex] using the formula rearranged as:

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

Substitute the values:

[tex]$$
h = \frac{144}{64} = 2.25
$$[/tex]

Thus, the hammer was dropped from a height of [tex]$\boxed{2.25\text{ feet}}$[/tex].

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Rewritten by : Barada