College

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

Answer :

Sure! Let's go through the problem together step by step.

We are given the following information:
1. The hammer hits the ground with a speed ([tex]\(v\)[/tex]) of 4 feet per second.
2. The acceleration due to gravity ([tex]\(g\)[/tex]) is 32 feet per second squared.

We need to find how far above the ground ([tex]\(h\)[/tex]) the hammer was when you dropped it. We'll use the equation:

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

We need to solve for [tex]\(h\)[/tex]. Let's rearrange the formula to solve for [tex]\(h\)[/tex]:

1. Square both sides of the equation 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]

Now, let's plug in the given values:

- [tex]\(v = 4\)[/tex] feet per second
- [tex]\(g = 32\)[/tex] feet per second squared

Substitute these into the equation:

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

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

[tex]\[ h = \frac{1}{4} \][/tex]

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

So, the hammer was 0.25 feet above the ground when it was dropped. Therefore, the correct answer is:

C. 0.25 feet

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