To solve this problem, we can use the following kinematic equation for uniformly accelerated motion (assuming downward is negative):

$d = v_0 t + \frac{1}{2} a t^2$

Where:

• $d$ is the displacement (height of the cliff),
• $v_0$ is the initial velocity ($-7.1 \, \text{m/s}$),
• $t$ is the time (7.5 s),
• $a$ is the acceleration due to gravity ($-9.8 \, \text{m/s}^2$).

Now, plug the values into the equation:

$d = (-7.1)(7.5) + \frac{1}{2} (-9.8)(7.5)^2$

Let's calculate this step by step.The height of the cliff is approximately 328.88 meters.

This negative value indicates that the displacement is downward, which is consistent with the rock being thrown downward. Therefore, the cliff's height is 328.88 meters.

Would you like more details, or do you have any questions?

Here are 5 related questions:

1. How would the result change if the initial velocity was zero?
2. What if the rock was thrown upward instead?
3. How would the presence of air resistance affect the outcome?
4. Can you calculate the final velocity of the rock when it hits the ground?
5. How long would the rock take if it fell freely from rest?

Tip: Always check the signs of velocities and accelerations when solving kinematics problems!