Answer :
To find out how far above the ground the hammer was initially dropped, we can use the formula for the speed of an object in free fall:
[tex]\[ v = \sqrt{2gh} \][/tex]
Where:
- [tex]\( v \)[/tex] is the final speed (4 feet per second),
- [tex]\( g \)[/tex] is the acceleration due to gravity (32 feet/second²),
- [tex]\( h \)[/tex] is the height above the ground.
We need to solve for [tex]\( h \)[/tex]. To do that, follow these steps:
1. Square both sides of the equation to eliminate the square root:
[tex]\[ v^2 = 2gh \][/tex]
2. Substitute the given values into the equation:
[tex]\[ 4^2 = 2 \times 32 \times h \][/tex]
3. Calculate [tex]\( v^2 \)[/tex]:
[tex]\[ 4^2 = 16 \][/tex]
4. Set up the equation with the known values:
[tex]\[ 16 = 64h \][/tex]
5. Solve for [tex]\( h \)[/tex]:
[tex]\[ h = \frac{16}{64} \][/tex]
6. Simplify the fraction:
[tex]\[ h = 0.25 \][/tex]
So, the hammer was dropped from a height of 0.25 feet above the ground. Therefore, the correct answer is:
B. 0.25 feet
[tex]\[ v = \sqrt{2gh} \][/tex]
Where:
- [tex]\( v \)[/tex] is the final speed (4 feet per second),
- [tex]\( g \)[/tex] is the acceleration due to gravity (32 feet/second²),
- [tex]\( h \)[/tex] is the height above the ground.
We need to solve for [tex]\( h \)[/tex]. To do that, follow these steps:
1. Square both sides of the equation to eliminate the square root:
[tex]\[ v^2 = 2gh \][/tex]
2. Substitute the given values into the equation:
[tex]\[ 4^2 = 2 \times 32 \times h \][/tex]
3. Calculate [tex]\( v^2 \)[/tex]:
[tex]\[ 4^2 = 16 \][/tex]
4. Set up the equation with the known values:
[tex]\[ 16 = 64h \][/tex]
5. Solve for [tex]\( h \)[/tex]:
[tex]\[ h = \frac{16}{64} \][/tex]
6. Simplify the fraction:
[tex]\[ h = 0.25 \][/tex]
So, the hammer was dropped from a height of 0.25 feet above the ground. Therefore, the correct answer is:
B. 0.25 feet
