To solve this problem, we can use the kinetic energy and gravitational potential energy relationship. We know that at the maximum height, all the rock's kinetic energy will have been converted into potential energy, so we can apply the work-energy principle. However, since we're working with the kinematic equations for projectile motion, a simpler approach is to use the following kinematic formula:

$v^2 = u^2 + 2a s$

Where:

• $v$ is the final velocity at the maximum height (which is 0 m/s, because the rock stops momentarily at the highest point),
• $u$ is the initial velocity (9 m/s),
• $a$ is the acceleration due to gravity (approximately $-9.8 \, \text{m/s}^2$),
• $s$ is the displacement (the maximum height increase).

Rearranging the equation to solve for $s$:

$0 = u^2 + 2a s$ $s = \frac{-u^2}{2a}$

Substituting the values:

$s = \frac{-(9)^2}{2(-9.8)}$ $s = \frac{-81}{-19.6}$ $s = 4.13 \, \text{m}$

Now, we need to account for the initial height from which the rock was thrown. Since the rock was thrown from a height of 1.5 meters, the total maximum height $H_{\text{total}}$ is:

$H_{\text{total}} = 1.5 + 4.13 = 5.63 \, \text{m}$

Thus, the maximum height the rock will reach is 5.63 meters.

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Would you like a more detailed explanation of the concepts or any further clarification? Here are some related questions:

1. How does the acceleration due to gravity affect the motion of a projectile?
2. What would happen to the maximum height if the initial velocity were doubled?
3. How can we calculate the time it takes for the rock to reach the maximum height?
4. How can the velocity of the rock be determined at various points during its flight?
5. What happens if the rock is thrown downward instead of upward?

Tip: Always remember that the maximum height occurs when the velocity is momentarily zero before the object starts descending.