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.
[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.
