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 [tex]^2[/tex], how far above the ground [tex](h)[/tex] was the hammer when you dropped it? Use the formula:



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



A. 0.25 feet

B. 0.5 feet

C. 1.0 foot

D. 16.0 feet

Answer :

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

### Explanation
1. Understanding the Problem
We are given the formula $v =
\sqrt{2gh}$, where $v$ is the speed of the hammer when it 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 $h$.

2. Squaring Both Sides
First, let's square both sides of the equation to eliminate the square root:$$v^2 = 2gh$$

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

4. Substituting Values
Substitute the given values $v = 4$ and $g = 32$ into the equation:$$h = \frac{4^2}{2 \times 32} = \frac{16}{64} = \frac{1}{4} = 0.25$$

5. Final Answer
Therefore, the height $h$ is 0.25 feet. The correct answer is A.

### Examples
Understanding the relationship between speed, gravity, and height is useful in various real-world scenarios. For example, engineers use these principles to design safety equipment, such as helmets or padding, to minimize the impact force during falls. Similarly, understanding projectile motion helps athletes optimize their performance in sports like basketball or baseball, where calculating the trajectory of a ball is crucial. This concept also applies to amusement park rides, where designers need to calculate the height and speed of drops to ensure a thrilling yet safe experience.