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The problem involves determining the height of a cliff from which a rock is thrown downward with an initial velocity of \( v_i = 7.50 \, \text{m/s} \). The time it takes to hit the ground is \( t = 4.50 \, \text{seconds} \), and we are given the kinematic equation:

\[
\Delta d = v_i t + \frac{1}{2} g t^2
\]

Where:
- \( \Delta d \) is the height of the cliff,
- \( v_i = 7.50 \, \text{m/s} \) is the initial velocity,
- \( g = 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity,
- \( t = 4.50 \, \text{seconds} \) is the time.

We will now calculate the height \( \Delta d \).

\[
\Delta d = 7.50 \times 4.50 + \frac{1}{2} \times 9.81 \times (4.50)^2
\]

Let me compute that.The height of the cliff is approximately \( 133 \, \text{m} \).

This matches option D in the given multiple-choice answers.

Would you like more details or have any further questions?

Here are 5 related questions:
1. How would the height of the cliff change if the initial velocity were 0 m/s?
2. What would happen if the rock were thrown upwards instead of downwards?
3. How does air resistance affect the time taken for the rock to reach the ground?
4. How does the equation of motion change if the rock were in a vacuum?
5. What is the significance of the time squared term in the equation?

**Tip**: Remember that when dealing with projectile motion, initial velocity and time are key factors that determine the distance an object travels.

Your friend throws a rock off of a cliff straight down with an initial speed of 7.50 m/s. Using a stopwatch, you determine it takes the rock 4.50 s to hit the ground below. How high is the cliff?

Learn how to calculate the height of a cliff when a rock is thrown straight down with an initial velocity of 7.50 m/s and hits the ground after 4.50 seconds. This problem uses the kinematic equation Δd = v_i * t + 1/2 * g * t^2 to find the height, applying concepts from projectile motion and gravity.

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