College

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

Answer :

Certainly! Let's solve the problem step-by-step using the given formula:

The formula you need is:

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

where:
- [tex]\( v \)[/tex] is the final speed when the hammer hits the floor, which is 12 feet per second.
- [tex]\( g \)[/tex] is the acceleration due to gravity, which is 32 feet/second².
- [tex]\( h \)[/tex] is the height from which the hammer was dropped.

We need to solve for [tex]\( h \)[/tex]. Start by rearranging the formula to solve for [tex]\( h \)[/tex]:

1. Square both sides to eliminate the square root:
[tex]\[ v^2 = 2gh \][/tex]

2. Solve for [tex]\( h \)[/tex] by dividing both sides by [tex]\( 2g \)[/tex]:
[tex]\[ h = \frac{v^2}{2g} \][/tex]

Insert the known values into the equation:

- [tex]\( v = 12 \)[/tex]
- [tex]\( g = 32 \)[/tex]

Now substitute those values into the equation:

[tex]\[ h = \frac{12^2}{2 \times 32} \][/tex]

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

When you simplify the fraction:

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

So, the hammer was dropped from a height of 2.25 feet. Therefore, the correct answer is:

A. 2.25 feet