High School

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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]

A. 16.0 feet
B. 1.0 foot
C. 2.0 feet
D. 8.0 feet

Answer :

We start with the formula for the speed of an object falling from a height:

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

Here,
- [tex]$v$[/tex] is the final speed,
- [tex]$g$[/tex] is the acceleration due to gravity, and
- [tex]$h$[/tex] is the height from which the object was dropped.

Since we want to find [tex]$h$[/tex], we first square both sides of the equation:

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

Then, we solve for [tex]$h$[/tex] by dividing both sides by [tex]$2g$[/tex]:

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

Given:
- [tex]$v = 8$[/tex] ft/s,
- [tex]$g = 32$[/tex] ft/s[tex]$^2$[/tex],

we substitute these values into the equation:

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

Calculating the numerator:

[tex]$$
8^2 = 64
$$[/tex]

And the denominator:

[tex]$$
2 \times 32 = 64
$$[/tex]

Thus, we have:

[tex]$$
h = \frac{64}{64} = 1 \text{ foot}
$$[/tex]

Therefore, the height above the ground when the hammer was dropped is [tex]$\boxed{1.0}$[/tex] foot.

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Rewritten by : Barada