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

Answer :

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

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

Here:
- [tex]\( v \)[/tex] is the final velocity of the hammer, which is 8 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, which we need to find.

First, we need to rearrange the formula to solve for [tex]\( h \)[/tex]. Start by squaring both sides of the equation to get rid of the square root:

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

Next, divide both sides by [tex]\( 2g \)[/tex] to isolate [tex]\( h \)[/tex]:

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

Now, substitute the known values into the equation:

- [tex]\( v = 8 \)[/tex] feet per second
- [tex]\( g = 32 \)[/tex] feet/second²

[tex]\[ h = \frac{8^2}{2 \times 32} \][/tex]

This simplifies to:

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

So, the height [tex]\( h \)[/tex] is:

[tex]\[ h = 1 \][/tex]

Therefore, the hammer was dropped from a height of 1 foot above the ground.

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Rewritten by : Barada