To model the height of a ball dropped from a height of 3 feet, we can use the physics of free fall. The height $h(t)$ of the ball as a function of time $t$ (in seconds) can be expressed with the equation:

$h(t) = h_0 - \frac{1}{2} g t^2$

where:

• $h_0$ is the initial height (3 feet),
• $g$ is the acceleration due to gravity (approximately $32 \, \text{ft/s}^2$),
• $t$ is the time in seconds.

Plugging in the values:

$h(t) = 3 - 16t^2$

To find how long the ball is in the air, we set the height $h(t)$ to 0 (when the ball hits the ground):

$0 = 3 - 16t^2$

Solving for $t$:

$16t^2 = 3$ $t^2 = \frac{3}{16}$ $t = \sqrt{\frac{3}{16}} = \frac{\sqrt{3}}{4}$

Calculating $\frac{\sqrt{3}}{4}$:

$t \approx 0.433 \, \text{seconds}$

Rounding to the nearest hundredths:

The ball was in the air for approximately 0.43 seconds.

Summary

• Function: $h(t) = 3 - 16t^2$
• Time in air: Approximately 0.43 seconds

Would you like to explore more about this topic or have any questions? Here are some related questions you might consider:

1. What factors can affect the time a ball stays in the air?
2. How does the height from which the ball is dropped influence its fall time?
3. What would the equation look like if the ball was dropped from a greater height?
4. How does air resistance affect a ball dropped from the same height?
5. Can you calculate the impact velocity of the ball just before it hits the ground?

Tip: When modeling free fall, always consider air resistance in real-life scenarios, as it can significantly alter the results compared to ideal conditions.