The formula $t = h \sqrt{4}$ doesn't seem correct for modeling the time it takes an object to fall. In physics, the time $t$ it takes for an object to fall from a height $h$ under the influence of gravity is typically given by the formula:

$t = \sqrt{\frac{2h}{g}}$

where $g$ is the acceleration due to gravity, approximately $32 \, \text{ft/s}^2$ (for objects falling near Earth's surface), and $h$ is the height in feet.

Given $h = 125 \, \text{feet}$, we can use this formula to estimate the time $t$.

Step-by-step solution:

1. The formula becomes: $t = \sqrt{\frac{2 \times 125}{32}}$
2. Simplifying inside the square root: $t = \sqrt{\frac{250}{32}} = \sqrt{7.8125}$
3. Estimate $\sqrt{7.8125}$ to the nearest integer: $\sqrt{7.8125} \approx 3$
4. So, the time it takes for the rock to fall is about $3$ seconds.

Thus, it will take approximately 3 seconds for the rock to hit the ground.

Would you like further details or have any questions?

Here are 5 related questions for exploration:

1. How does the height $h$ affect the time $t$ for an object to fall?
2. What is the significance of the constant $g = 32 \, \text{ft/s}^2$?
3. How would the time change if the object was dropped from a height of 250 feet?
4. How does air resistance affect the time an object takes to fall?
5. Can we use the same formula for an object falling on the Moon?

Tip: Estimating square roots to the nearest integer is useful for quick approximations in physics problems.