Answer :
To determine how far above the ground the hammer was when it was dropped, we use the formula:
[tex]\[ v = \sqrt{2gh} \][/tex]
where:
- [tex]\( v \)[/tex] is the speed at which the hammer hits the ground (4 feet per second),
- [tex]\( g \)[/tex] is the acceleration due to gravity (32 feet per second squared),
- [tex]\( h \)[/tex] is the height above the ground.
We want to find [tex]\( h \)[/tex], so we'll rearrange the formula:
1. First, square both sides to eliminate the square root:
[tex]\[ v^2 = 2gh \][/tex]
2. Now, solve for [tex]\( h \)[/tex]:
[tex]\[ h = \frac{v^2}{2g} \][/tex]
Substitute the given values into the equation:
- [tex]\( v = 4 \)[/tex] feet per second
- [tex]\( g = 32 \)[/tex] feet per second squared
Calculate [tex]\( h \)[/tex]:
[tex]\[ h = \frac{(4)^2}{2 \times 32} \][/tex]
[tex]\[ h = \frac{16}{64} \][/tex]
[tex]\[ h = 0.25 \text{ feet} \][/tex]
Therefore, the hammer was dropped from a height of 0.25 feet above the ground. The correct answer is C. 0.25 feet.
[tex]\[ v = \sqrt{2gh} \][/tex]
where:
- [tex]\( v \)[/tex] is the speed at which the hammer hits the ground (4 feet per second),
- [tex]\( g \)[/tex] is the acceleration due to gravity (32 feet per second squared),
- [tex]\( h \)[/tex] is the height above the ground.
We want to find [tex]\( h \)[/tex], so we'll rearrange the formula:
1. First, square both sides to eliminate the square root:
[tex]\[ v^2 = 2gh \][/tex]
2. Now, solve for [tex]\( h \)[/tex]:
[tex]\[ h = \frac{v^2}{2g} \][/tex]
Substitute the given values into the equation:
- [tex]\( v = 4 \)[/tex] feet per second
- [tex]\( g = 32 \)[/tex] feet per second squared
Calculate [tex]\( h \)[/tex]:
[tex]\[ h = \frac{(4)^2}{2 \times 32} \][/tex]
[tex]\[ h = \frac{16}{64} \][/tex]
[tex]\[ h = 0.25 \text{ feet} \][/tex]
Therefore, the hammer was dropped from a height of 0.25 feet above the ground. The correct answer is C. 0.25 feet.
