Let's solve this problem by using the equation of motion for the position of the rock as a function of time. The formula for the height $h(t)$ of the rock at any time $t$ is given by:

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

where:

• $h_0 = 20$ m is the initial height of the rock.
• $v_0 = 24$ m/s is the initial velocity of the rock.
• $g = 9.8$ m/s$^2$ is the acceleration due to gravity.
• $t$ is the time in seconds.

We want to find the times $t$ when the rock is 7 meters above the ground, so we set $h(t) = 7$ and solve for $t$.

$7 = 20 + 24t - \frac{1}{2} \times 9.8 \times t^2$

Simplifying this equation:

$7 = 20 + 24t - 4.9t^2$

$0 = 4.9t^2 - 24t + 13$

Now, let's solve this quadratic equation for $t$. We'll use the quadratic formula:

$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = 4.9$, $b = -24$, and $c = 13$. Let's calculate the values of $t$.The solutions for the time $t$ when the rock is 7 meters above the ground are:

$t_1 = 4.28 \, \text{seconds}$ $t_2 = 0.62 \, \text{seconds}$

These times correspond to the moments when the rock is rising to 7 meters (0.62 seconds) and when it is descending past 7 meters (4.28 seconds).

Would you like to go into more detail or have any questions?

Here are some related questions:

1. How would the times change if the initial velocity were different?
2. What would happen if the cliff height was increased?
3. How does changing the acceleration due to gravity affect the results?
4. What are the maximum height and the time at which it occurs?
5. How would the equation change if the rock were thrown downward?

Tip: When solving motion problems, carefully consider the direction of forces and velocities.