High School

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{2 g h} [/tex]

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

Answer :

To find out how far above the ground the hammer was initially dropped, we can use the formula for the speed of an object in free fall:

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

Where:
- [tex]\( v \)[/tex] is the final speed (4 feet per second),
- [tex]\( g \)[/tex] is the acceleration due to gravity (32 feet/second²),
- [tex]\( h \)[/tex] is the height above the ground.

We need to solve for [tex]\( h \)[/tex]. To do that, follow these steps:

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

[tex]\[ v^2 = 2gh \][/tex]

2. Substitute the given values into the equation:

[tex]\[ 4^2 = 2 \times 32 \times h \][/tex]

3. Calculate [tex]\( v^2 \)[/tex]:

[tex]\[ 4^2 = 16 \][/tex]

4. Set up the equation with the known values:

[tex]\[ 16 = 64h \][/tex]

5. Solve for [tex]\( h \)[/tex]:

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

6. Simplify the fraction:

[tex]\[ h = 0.25 \][/tex]

So, the hammer was dropped from a height of 0.25 feet above the ground. Therefore, the correct answer is:

B. 0.25 feet