To find the velocity of the rock after 1 second, we need to calculate the derivative of the height function $H(t) = 17t - 1.86t^2$ with respect to time $t$. The derivative of the height function represents the velocity.

The velocity function $v(t)$ is given by:

$v(t) = \frac{dH}{dt} = \frac{d}{dt} \left( 17t - 1.86t^2 \right)$

Differentiating:

$v(t) = 17 - 2(1.86)t = 17 - 3.72t$

Now, substitute $t = 1$ second to find the velocity after 1 second:

$v(1) = 17 - 3.72(1) = 17 - 3.72 = 13.28 \, \text{m/s}$

Thus, the velocity of the rock after 1 second is 13.28 m/s.

Would you like further details or clarification? Here are 5 related questions to extend your understanding:

1. What is the velocity at the instant the rock is released (at $t = 0$)?
2. How would you find the time at which the rock reaches its maximum height?
3. What is the significance of the negative coefficient in the $t^2$ term for the height function?
4. Can you calculate the velocity at $t = 2$ seconds?
5. How would you determine the total time the rock spends in the air before hitting the ground?

Tip: The derivative of a position function with respect to time always gives the velocity. Keep this in mind for motion problems!