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Let's analyze the given problem and solve each part step-by-step.

The height \( s \) of the rock at time \( t \) is given by:
\[
s = 32t - 0.8t^2
\]

### Part (a): Find the rock's velocity and acceleration at time \( t \).
The velocity \( v \) is the first derivative of the height \( s \) with respect to time \( t \):
\[
v = \frac{ds}{dt} = 32 - 1.6t
\]

The acceleration \( a \) is the derivative of the velocity \( v \) with respect to \( t \):
\[
a = \frac{dv}{dt} = -1.6 \text{ m/s}^2
\]

So:
- **Velocity**: \( v = 32 - 1.6t \)
- **Acceleration**: \( a = -1.6 \, \text{m/s}^2 \)

### Part (b): How long does it take the rock to reach its highest point?
The rock reaches its highest point when the velocity is zero, i.e., \( v = 0 \):
\[
32 - 1.6t = 0
\]
\[
1.6t = 32
\]
\[
t = \frac{32}{1.6} = 20 \text{ seconds}
\]

So, it takes **20 seconds** to reach the highest point.

### Part (c): How high does the rock go?
To find the maximum height, substitute \( t = 20 \) seconds into the height equation:
\[
s = 32(20) - 0.8(20)^2
\]
\[
s = 640 - 0.8 \times 400
\]
\[
s = 640 - 320 = 320 \text{ meters}
\]

The maximum height reached by the rock is **320 meters**.

### Part (d): How long does it take the rock to reach half its maximum height?
Half of the maximum height is \( \frac{320}{2} = 160 \) meters. We set \( s = 160 \) and solve for \( t \):
\[
160 = 32t - 0.8t^2
\]
Rearrange this into standard quadratic form:
\[
0.8t^2 - 32t + 160 = 0
\]
Divide by \( 0.8 \):
\[
t^2 - 40t + 200 = 0
\]
Now, use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[
t = \frac{40 \pm \sqrt{(-40)^2 - 4 \cdot 1 \cdot 200}}{2 \cdot 1}
\]
\[
t = \frac{40 \pm \sqrt{1600 - 800}}{2}
\]
\[
t = \frac{40 \pm \sqrt{800}}{2}
\]
\[
t = \frac{40 \pm 20\sqrt{2}}{2}
\]
\[
t = 20 \pm 10\sqrt{2}
\]

Calculating \( 10\sqrt{2} \approx 14.14 \):
- \( t = 20 + 14.14 = 34.14 \) seconds (on the way down)
- \( t = 20 - 14.14 = 5.86 \) seconds (on the way up)

Therefore, it takes approximately **5.86 seconds** to reach half the maximum height on the way up.

### Part (e): How long is the rock aloft?
The rock is aloft until it returns to the ground, where \( s = 0 \):
\[
0 = 32t - 0.8t^2
\]
\[
0.8t^2 = 32t
\]
\[
t(0.8t - 32) = 0
\]
This gives \( t = 0 \) or \( 0.8t = 32 \):
\[
t = \frac{32}{0.8} = 40 \text{ seconds}
\]

Thus, the rock is aloft for **40 seconds**.

---

Would you like further details on any of these steps, or have any questions? 

### Related Questions:
1. How would the problem change if the gravitational acceleration on the Moon was different?
2. Can you derive the total displacement of the rock during its flight?
3. What would be the maximum height if the initial velocity was doubled?
4. How would air resistance affect the rock's time aloft?
5. Could you solve this using energy principles instead of kinematics?

#### Tip:
For projectile motion questions like this, always start by finding the velocity and acceleration expressions first; they will guide you to solve other parts of the problem more effectively.

A rock thrown vertically upward from the surface of the moon at a velocity of 32 m/sec reaches a height of s = 32t - 0.8t^2 meters in t sec. Solve for velocity, acceleration, time to highest point, maximum height, half-height time, and time aloft.

This detailed solution shows how to find the velocity, acceleration, maximum height, and time aloft for a rock thrown upward on the moon with an initial velocity of 32 m/s. Using differentiation and quadratic equations, you can solve for key values such as the rock's time to reach its peak and the total flight duration.

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