To determine the displacement of the rock dropped by Peppy the chimp, we can use the kinematic equation for free fall, assuming there's no air resistance:

$d = \frac{1}{2} g t^2$

Where:

• $d$ is the displacement (how far the rock falls),
• $g$ is the acceleration due to gravity ($9.8 \, \text{m/s}^2$),
• $t$ is the time it takes to fall (2 seconds).

Now, substituting the values:

$d = \frac{1}{2} \times 9.8 \, \text{m/s}^2 \times (2 \, \text{s})^2$ $d = \frac{1}{2} \times 9.8 \, \text{m/s}^2 \times 4 \, \text{s}^2$ $d = \frac{1}{2} \times 39.2 \, \text{m}$ $d = 19.6 \, \text{m}$

Thus, the displacement of the rock is 19.6 meters.

Would you like more details or have any other questions?

Here are 5 related questions:

1. What happens to the speed of the rock as it falls deeper?
2. How would the displacement change if Peppy dropped the rock on the Moon?
3. What is the velocity of the rock just before it hits the bottom?
4. How would air resistance affect the rock’s fall?
5. How can we calculate the time taken for the sound of the rock hitting the bottom to return to Peppy?

Tip: When dealing with free fall, the key equation to remember is $d = \frac{1}{2} g t^2$, which relates time, gravity, and displacement.