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The problem involves an apple falling from a tree, and we need to determine how high the apple was before it fell. The following details are provided:

- Acceleration due to gravity (\(a\)) = \(-9.8 \, \text{m/s}^2\)
- Initial velocity (\(v_i\)) = 0 m/s (since the apple starts from rest)
- Time of fall (\(t\)) = 1.92 seconds
- The unknown: distance (\(d\)).

### Step-by-step solution:
We use the kinematic equation:
\[
d = v_i t + \frac{1}{2} a t^2
\]
Given:
- \(v_i = 0\)
- \(a = -9.8 \, \text{m/s}^2\)
- \(t = 1.92 \, \text{seconds}\)

Substitute these values into the equation:
\[
d = 0 \times 1.92 + \frac{1}{2} \times (-9.8) \times (1.92)^2
\]
\[
d = 0 + 0.5 \times (-9.8) \times (1.92)^2
\]

Let me calculate the result.The height from which the apple fell is approximately \(18.06 \, \text{m}\). Since the distance is negative in our calculation, it reflects the downward direction of the fall.

Would you like further details or clarifications? 

Here are 5 related questions to expand your understanding:
1. How would the height change if the apple fell for a longer time?
2. What if the initial velocity wasn't zero? How would the equation change?
3. How do we account for air resistance in real-world scenarios?
4. Can you derive other kinematic equations for uniformly accelerated motion?
5. What is the difference between free fall and motion with constant acceleration?

**Tip:** Always remember that a negative sign in physics can indicate direction, especially in kinematics involving free fall.

An apple falls from the tree, and the fall is 1.92 seconds. How high up was the apple? (given g = -9.8 m/s²)

Learn how to calculate the height from which an apple fell in a free fall motion using the kinematic equation d = v_i * t + (1/2) * a * t^2. The initial velocity is 0 m/s, acceleration due to gravity is -9.8 m/s², and time of fall is 1.92 seconds. Suitable for high school physics students.

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