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Answer :
We start with the formula for the speed of an object falling under gravity:
[tex]$$
v = \sqrt{2gh}
$$[/tex]
where:
- [tex]$v$[/tex] is the speed when striking the floor,
- [tex]$g$[/tex] is the acceleration due to gravity, and
- [tex]$h$[/tex] is the height from which the object was dropped.
Given:
- [tex]$v = 8$[/tex] ft/s,
- [tex]$g = 32$[/tex] ft/s[tex]$^2$[/tex].
First, we square both sides of the equation to eliminate the square root:
[tex]$$
v^2 = 2gh
$$[/tex]
Now, solve for [tex]$h$[/tex] by dividing both sides by [tex]$2g$[/tex]:
[tex]$$
h = \frac{v^2}{2g}
$$[/tex]
Substitute the given values into the equation:
[tex]$$
h = \frac{8^2}{2 \times 32} = \frac{64}{64} = 1.0
$$[/tex]
Thus, the hammer was dropped from a height of [tex]$1.0$[/tex] foot above the ground.
The correct answer is: [tex]$\boxed{1.0\ \text{foot}}$[/tex].
[tex]$$
v = \sqrt{2gh}
$$[/tex]
where:
- [tex]$v$[/tex] is the speed when striking the floor,
- [tex]$g$[/tex] is the acceleration due to gravity, and
- [tex]$h$[/tex] is the height from which the object was dropped.
Given:
- [tex]$v = 8$[/tex] ft/s,
- [tex]$g = 32$[/tex] ft/s[tex]$^2$[/tex].
First, we square both sides of the equation to eliminate the square root:
[tex]$$
v^2 = 2gh
$$[/tex]
Now, solve for [tex]$h$[/tex] by dividing both sides by [tex]$2g$[/tex]:
[tex]$$
h = \frac{v^2}{2g}
$$[/tex]
Substitute the given values into the equation:
[tex]$$
h = \frac{8^2}{2 \times 32} = \frac{64}{64} = 1.0
$$[/tex]
Thus, the hammer was dropped from a height of [tex]$1.0$[/tex] foot above the ground.
The correct answer is: [tex]$\boxed{1.0\ \text{foot}}$[/tex].
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