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

Answer :

To find out how far above the ground the hammer was when it was dropped, we will use the formula given:

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

where:
- [tex]\( v \)[/tex] is the velocity of the hammer when it hits the floor, which is 4 feet per second.
- [tex]\( g \)[/tex] is the acceleration due to gravity, which is 32 feet/second².
- [tex]\( h \)[/tex] is the height above the ground from which the hammer was dropped.

Our goal is to solve for [tex]\( h \)[/tex]. Let's go through the steps:

1. Start with the formula:

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

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

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

3. Substitute the given values into the equation. We have [tex]\( v = 4 \)[/tex] and [tex]\( g = 32 \)[/tex]:

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

4. Calculate [tex]\( 4^2 \)[/tex] which is 16:

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

5. Solve for [tex]\( h \)[/tex] by dividing both sides by 64:

[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 option C: 0.25 feet.