Answer :
The problem involves finding the initial height of a dropped hammer given its final speed and the acceleration due to gravity. The formula $v =
\sqrt{2gh}$ is used.
- Square both sides: $v^2 = 2gh$.
- Solve for $h$: $h = \frac{v^2}{2g}$.
- Substitute $v = 8$ and $g = 32$: $h = \frac{8^2}{2 \times 32} = 1$.
- The initial height is $\boxed{1 \text{ foot}}$.
### Explanation
1. Understanding the Problem
We are given the formula $v =
\sqrt{2gh}$, where $v$ is the final velocity of the hammer, $g$ is the acceleration due to gravity, and $h$ is the initial height. We are given that $v = 8$ feet per second and $g = 32$ feet/second${}^2$. We want to find the height $h$.
2. Squaring Both Sides
First, let's square both sides of the equation to get rid of the square root:
$v^2 = 2gh$.
3. Isolating h
Now, we want to solve for $h$, so we divide both sides by $2g$:
$h = \frac{v^2}{2g}$.
4. Plugging in Values
Now, we plug in the given values for $v$ and $g$:
$h = \frac{8^2}{2 \times 32} = \frac{64}{64} = 1$.
5. Final Answer
So, the initial height of the hammer above the ground was 1 foot.
### Examples
Understanding the relationship between the height from which an object is dropped and its impact speed is crucial in various real-world scenarios, such as designing safety equipment, analyzing the impact of falling objects, and understanding projectile motion in sports. For instance, engineers use these principles to calculate the necessary padding in helmets or the impact force of debris falling from buildings, ensuring safety and minimizing potential damage.
\sqrt{2gh}$ is used.
- Square both sides: $v^2 = 2gh$.
- Solve for $h$: $h = \frac{v^2}{2g}$.
- Substitute $v = 8$ and $g = 32$: $h = \frac{8^2}{2 \times 32} = 1$.
- The initial height is $\boxed{1 \text{ foot}}$.
### Explanation
1. Understanding the Problem
We are given the formula $v =
\sqrt{2gh}$, where $v$ is the final velocity of the hammer, $g$ is the acceleration due to gravity, and $h$ is the initial height. We are given that $v = 8$ feet per second and $g = 32$ feet/second${}^2$. We want to find the height $h$.
2. Squaring Both Sides
First, let's square both sides of the equation to get rid of the square root:
$v^2 = 2gh$.
3. Isolating h
Now, we want to solve for $h$, so we divide both sides by $2g$:
$h = \frac{v^2}{2g}$.
4. Plugging in Values
Now, we plug in the given values for $v$ and $g$:
$h = \frac{8^2}{2 \times 32} = \frac{64}{64} = 1$.
5. Final Answer
So, the initial height of the hammer above the ground was 1 foot.
### Examples
Understanding the relationship between the height from which an object is dropped and its impact speed is crucial in various real-world scenarios, such as designing safety equipment, analyzing the impact of falling objects, and understanding projectile motion in sports. For instance, engineers use these principles to calculate the necessary padding in helmets or the impact force of debris falling from buildings, ensuring safety and minimizing potential damage.
