Assuming no air resistance or forces other than gravity, the height above the ground at time t of a falling object is given by es007-1.jpg where h is the height in meters, t is in seconds, v0 isthe object’s initial velocity in meters per second, and h0 is its initial (height) position in meters.

a) What is the applied domain of this function?
b) Give a scenario in which h0 <0.

For the given problem, the height \( h \) of the falling object at time \( t \) is determined by the formula:

\[
h(t) = h_0 + v_0 t - \frac{1}{2} g t^2
\]

where:
- \( h_0 \) is the initial height (in meters),
- \( v_0 \) is the initial velocity (in meters per second),
- \( g \) is the acceleration due to gravity (\( g \approx 9.8 \, \text{m/s}^2 \)) on Earth, and
- \( t \) is the time (in seconds) after the object begins to fall.

### a) **What is the applied domain of this function?**

The domain of the function represents all possible values of \( t \), the time variable. Since the object starts falling at time \( t = 0 \) and continues until it hits the ground (where \( h(t) = 0 \)), the domain will be limited by these physical constraints.

- \( t \geq 0 \): Time cannot be negative, as we're considering the motion after the object starts falling.
- The upper limit of \( t \) occurs when \( h(t) = 0 \), meaning when the object reaches the ground.

So, the domain is:

\[
t \in [0, t_{\text{ground}}]
\]

where \( t_{\text{ground}} \) is the time at which the object hits the ground, which can be found by solving \( h(t_{\text{ground}}) = 0 \).

### b) **Give a scenario in which \( h_0 < 0 \).**

A scenario where \( h_0 < 0 \) (the initial height is below ground level) could occur in the following situation:

Imagine a mine or tunnel below the ground. An object is dropped from a point inside this tunnel, which is beneath the Earth's surface (i.e., below the ground level). In this case, \( h_0 \) would be negative, indicating that the object starts falling from a position lower than the ground level, such as from a depth of 5 meters underground. If the object is released from that position, the initial height would be \( h_0 = -5 \, \text{m} \), which is below ground.

#### Example Scenario:
- A miner drops a tool from a depth of 10 meters below ground, so \( h_0 = -10 \, \text{m} \).

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Do you want further details or have any questions on this topic? Here are some related questions you might find useful:

1. How do you solve for \( t_{\text{ground}} \) when the object hits the ground?
2. How does the initial velocity \( v_0 \) affect the trajectory of the object?
3. What happens if the object is thrown upwards (i.e., \( v_0 > 0 \)) rather than dropped?
4. Can the height \( h(t) \) ever become negative during the motion?
5. How does the value of gravitational acceleration \( g \) change on different planets?

**Tip**: When solving quadratic equations like \( h(t) = 0 \), remember to consider both roots of the equation—only the positive root will be physically meaningful in this context.

Assuming no air resistance or forces other than gravity, the height above the ground at time t of a falling object is given by es007-1.jpg where h is the height in meters, t is in seconds, v0 is the object’s initial velocity in meters per second, and h0 is its initial (height) position in meters.

a) What is the applied domain of this function?
b) Give a scenario in which h0 < 0.

Explore the kinematics of a falling object with no air resistance. Learn how to apply the formula h(t) = h_0 + v_0t - (1/2)gt^2 to find the time it takes for an object to hit the ground and scenarios where the initial height h_0 is negative. Suitable for high school physics students in Grades 9-12.

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