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The problem you're working on is based on a rock being thrown upward on Mars, and the height \( H \) of the rock at time \( t \) is modeled by the equation:

\[
H = 17t - 1.86t^2
\]

Let's solve each part step by step:

### (a) Find the velocity of the rock after 2 seconds.

The velocity of the rock is the derivative of the height equation with respect to time \( t \). So, we differentiate \( H(t) \) to get \( v(t) \):

\[
v(t) = \frac{dH}{dt} = 17 - 2(1.86)t = 17 - 3.72t
\]

Now, substitute \( t = 2 \) seconds:

\[
v(2) = 17 - 3.72 \times 2 = 17 - 7.44 = 9.56 \, \text{m/s}
\]

So, the velocity after 2 seconds is **9.56 m/s**.

---

### (b) Find the velocity of the rock when \( t = a \).

For this part, just plug in \( t = a \) into the velocity equation:

\[
v(a) = 17 - 3.72a
\]

This is the general expression for the velocity when \( t = a \).

---

### (c) When will the rock hit the surface?

The rock hits the surface when the height \( H = 0 \). Set the height equation to zero and solve for \( t \):

\[
0 = 17t - 1.86t^2
\]

Factor the equation:

\[
t(17 - 1.86t) = 0
\]

So, \( t = 0 \) or \( 17 - 1.86t = 0 \). Solving for \( t \) when \( t \neq 0 \):

\[
1.86t = 17
\]
\[
t = \frac{17}{1.86} \approx 9.1 \, \text{seconds}
\]

So, the rock hits the surface after **9.1 seconds**.

---

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

To find the velocity when the rock hits the surface, substitute \( t = 9.1 \) seconds into the velocity equation:

\[
v(9.1) = 17 - 3.72 \times 9.1 = 17 - 33.852 = -16.85 \, \text{m/s}
\]

The negative sign indicates the rock is moving downward, so the velocity is **16.85 m/s** downward when it hits the surface.

---

Would you like a detailed explanation of any part? Here are five related questions to consider:

1. How would the results change if the initial velocity were different?
2. What does the negative velocity indicate about the direction of motion?
3. How does the gravity on Mars affect the time it takes for the rock to reach the ground compared to Earth?
4. Could we determine the maximum height of the rock, and if so, how?
5. What assumptions are made in this model about air resistance?

**Tip:** Derivatives help calculate the rate of change, which is why we used the derivative of height to get the velocity.

If a rock is thrown upward on the planet Mars with a velocity of 17 m/s, its height above the ground (in meters) after t seconds is given by H = 17t - 1.86t^2. (a) Find the velocity (in m/s) of the rock after 2 seconds. (b) Find the velocity (in m/s) of the rock when t = a. (c) When (in seconds) will the rock hit the surface? (d) With what velocity (in m/s) will the rock hit the surface?

Learn how to calculate the velocity of a rock thrown upward on Mars, find the time it hits the surface, and its final velocity. This involves using kinematics, derivatives, and quadratic equations. Suitable for high school students in Grades 11-12.

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