Answer :
We start with the formula for the speed when an object falls freely:
[tex]$$
v = \sqrt{2gh}
$$[/tex]
Given that the final speed is [tex]$v = 12$[/tex] feet per second and the acceleration due to gravity is [tex]$g = 32$[/tex] feet per second squared, we first square the speed:
[tex]$$
v^2 = 12^2 = 144
$$[/tex]
Next, multiply the gravitational acceleration by [tex]$2$[/tex]:
[tex]$$
2g = 2 \times 32 = 64
$$[/tex]
Now, solve for the height [tex]$h$[/tex] using the formula rearranged as:
[tex]$$
h = \frac{v^2}{2g}
$$[/tex]
Substitute the values:
[tex]$$
h = \frac{144}{64} = 2.25
$$[/tex]
Thus, the hammer was dropped from a height of [tex]$\boxed{2.25\text{ feet}}$[/tex].
[tex]$$
v = \sqrt{2gh}
$$[/tex]
Given that the final speed is [tex]$v = 12$[/tex] feet per second and the acceleration due to gravity is [tex]$g = 32$[/tex] feet per second squared, we first square the speed:
[tex]$$
v^2 = 12^2 = 144
$$[/tex]
Next, multiply the gravitational acceleration by [tex]$2$[/tex]:
[tex]$$
2g = 2 \times 32 = 64
$$[/tex]
Now, solve for the height [tex]$h$[/tex] using the formula rearranged as:
[tex]$$
h = \frac{v^2}{2g}
$$[/tex]
Substitute the values:
[tex]$$
h = \frac{144}{64} = 2.25
$$[/tex]
Thus, the hammer was dropped from a height of [tex]$\boxed{2.25\text{ feet}}$[/tex].
