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.
[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.
