College

You are helping with some repairs at home. You drop a hammer, and it hits the floor at a speed of 4 feet per second. If the acceleration due to gravity ([tex]g[/tex]) is 32 feet/second\(^2\), how far above the ground ([tex]h[/tex]) was the hammer when you dropped it? Use the formula:



\[ v = \sqrt{2gh} \]



A. 1.0 foot

B. 16.0 feet

C. 0.5 feet

D. 0.25 feet

Answer :

- Square both sides of the equation: $v^2 = 2gh$.
- Solve for $h$: $h = \frac{v^2}{2g}$.
- Substitute $v = 4$ and $g = 32$: $h = \frac{4^2}{2 \times 32} = \frac{16}{64}$.
- Simplify: $h = 0.25$. The height is $\boxed{0.25}$ feet

### Explanation
1. Understanding the Problem
We are given the formula $v =
\sqrt{2gh}$, where $v$ is the speed at which the hammer hits the ground, $g$ is the acceleration due to gravity, and $h$ is the height from which the hammer was dropped. We are given $v = 4$ feet per second and $g = 32$ feet/second$^2$. We want to find the height $h$.

2. Eliminating the Square Root
First, let's square both sides of the equation to eliminate the square root:
$v^2 = 2gh$.

3. Solving for h
Now, we solve for $h$ by dividing both sides by $2g$:
$h = \frac{v^2}{2g}$.

4. Substituting the Values
Substitute the given values of $v = 4$ and $g = 32$ into the equation:
$h = \frac{4^2}{2 \times 32} = \frac{16}{64}$.

5. Simplifying the Fraction
Simplify the fraction:
$h = \frac{16}{64} = \frac{1}{4} = 0.25$ feet.

6. Final Answer
Therefore, the hammer was dropped from a height of 0.25 feet.

### Examples
Understanding the relationship between speed, gravity, and height can be applied in various real-world scenarios, such as calculating the height from which an object fell based on its impact speed. This principle is used in forensic science to estimate the height of a fall in accident investigations. It's also relevant in engineering when designing safety measures for falling objects or in sports to analyze the trajectory and impact of projectiles.