We are given the formula

$t = \frac{1}{4} \sqrt{25 - h},$

where $t$ is the time (in seconds) and $h$ is the height (in feet) above the water. We are asked to find the height $h$ when $t = 1$ second.

Step 1: Substitute $t = 1$ into the equation.

$1 = \frac{1}{4} \sqrt{25 - h}.$

Step 2: Eliminate the fraction by multiplying through by 4.

$4 = \sqrt{25 - h}.$

Step 3: Square both sides to remove the square root.

$16 = 25 - h.$

Step 4: Solve for $h$.

$h = 25 - 16,$ $h = 9.$

Final Answer:

The ball will be 9 feet above the water after 1 second.

Would you like further details or have additional questions? Here are related questions you might explore:

1. What is the height of the ball after 2 seconds?
2. How long does it take for the ball to hit the water (when $h = 0$)?
3. Can you derive a function to find $t$ for any height $h$?
4. What is the maximum height achievable with this formula, if the ball were dropped from a higher point?
5. How does the formula relate to the motion of free-falling objects under gravity?

Tip: When solving equations with square roots, always double-check for potential extraneous solutions caused by squaring both sides.