To find the distance $h$ an object has fallen when its velocity $v$ is 224 feet per second, we can use the formula:

$v = \sqrt{2gh}$

Given:

• $v = 224$ ft/s
• $g = 32$ ft/s²

First, we square both sides of the equation to eliminate the square root:

$v^2 = 2gh$

Substituting the known values:

$(224)^2 = 2 \times 32 \times h$

Calculating $224^2$:

$224^2 = 50176$

Now, substituting this into the equation:

$50176 = 64h$

Next, we solve for $h$:

$h = \frac{50176}{64}$

Calculating $h$:

$h = 784 \text{ feet}$

Thus, the object has fallen 784 feet.

Would you like any further details or explanations? Here are some related questions you might consider:

1. How does changing the value of $g$ affect the height?
2. What would the velocity be if the object fell from 1000 feet?
3. How does this formula apply to different planets with varying gravity?
4. What are the implications of air resistance on falling objects?
5. Can this formula be used for objects thrown downward?

Tip: Always check the units when working with formulas to ensure consistency and accuracy.