The problem provides the height $s(t)$ of an arrow shot straight up on the moon, given as:

$s(t) = 58t - 0.83t^2$

where $s(t)$ is the height in meters at time $t$ seconds. Let's break down each question:

1. Velocity of the arrow after 10 seconds:

   The velocity is the derivative of the height function $s(t)$, so we first compute $v(t) = \frac{ds}{dt}$.

   $v(t) = \frac{d}{dt} \left( 58t - 0.83t^2 \right) = 58 - 2(0.83)t = 58 - 1.66t$

   After 10 seconds:

   $v(10) = 58 - 1.66 \times 10 = 58 - 16.6 = 41.4 \, \text{m/s}$

2. Time for the arrow to return and hit the moon:

   The arrow will return to the moon when its height $s(t) = 0$. Solve $58t - 0.83t^2 = 0$.

   Factor the equation:

   $t(58 - 0.83t) = 0$

   So, $t = 0$ (initial launch) or $58 - 0.83t = 0$.

   Solving $58 = 0.83t$:

   $t = \frac{58}{0.83} \approx 69.88 \, \text{seconds}$

3. Velocity at the moment the arrow hits the moon:

   The velocity at the time the arrow returns is calculated using $v(t) = 58 - 1.66t$ at $t = 69.88$:

   $v(69.88) = 58 - 1.66 \times 69.88 \approx 58 - 116 = -58 \, \text{m/s}$

   The negative sign indicates the arrow is traveling downward, and its magnitude is 58 m/s.

If you need more details or have questions, feel free to ask!

Here are some related questions to expand understanding:

1. What is the significance of the negative velocity when the arrow hits the moon?
2. How does the quadratic term $0.83t^2$ affect the motion of the arrow?
3. How would the trajectory change if the initial velocity were different?
4. How do air resistance and gravity on Earth differ from this scenario on the moon?
5. What would the maximum height of the arrow be, and how can we calculate it?

Tip: The time to reach maximum height is when the velocity $v(t) = 0$.