Let's go through the problem and solve it step by step based on the given equation and tasks.

The height $H$ of a rock thrown upward is modeled by the equation:

$H(t) = 11t - 1.86t^2$

Where:

• $t$ is time in seconds,
• The velocity of the rock is the derivative of the height equation with respect to $t$: $v(t) = \frac{dH}{dt}$.

(a) Velocity after 1 second:

First, let's differentiate the height equation:

$v(t) = \frac{d}{dt} \left( 11t - 1.86t^2 \right) = 11 - 2(1.86)t = 11 - 3.72t$

At $t = 1$ second:

$v(1) = 11 - 3.72(1) = 11 - 3.72 = 7.28 \, \text{m/s}$

(b) Find the velocity when $t = a$:

If $t = a$, the velocity is:

$v(a) = 11 - 3.72a$

You would need to substitute the value of $a$ once it's known.

(c) When will the rock hit the surface? (i.e., when $H(t) = 0$):

Set the height equation equal to zero:

$0 = 11t - 1.86t^2$

Factoring the equation:

$t(11 - 1.86t) = 0$

This gives two solutions:

$t = 0 \quad \text{(the initial moment of throw)}$

$11 - 1.86t = 0 \implies t = \frac{11}{1.86} \approx 5.91 \, \text{seconds}$

So, the rock will hit the surface after approximately 5.9 seconds.

(d) With what velocity will the rock hit the surface?

To find the velocity when the rock hits the surface, substitute $t = 5.91$ into the velocity equation:

$v(5.91) = 11 - 3.72(5.91) \approx 11 - 22.0 = -11 \, \text{m/s}$

The negative sign indicates the rock is moving downward when it hits the surface, so the speed is 11 m/s downward.

Let me know if you'd like further clarification or more details.

Here are five related questions:

1. How would the equation change if the initial velocity were different?
2. What is the significance of the negative velocity?
3. How would the time of flight change if the gravity on Mars were stronger?
4. Can we calculate the maximum height the rock reaches?
5. How does the shape of the height function help visualize the motion?

Tip: Remember, the velocity equation gives insight into how fast and in what direction the object is moving at any point in time.