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 − 1.86t2.
(a)
Find the velocity (in m/s) of the rock after 2 seconds.
m/s
(b)
Find the velocity (in m/s) of the rock when 
t = a.
 m/s
(c)
When (in seconds) will the rock hit the surface? (Round your answer to one decimal place.)
t = 
 s
(d)
With what velocity (in m/s) will the rock hit the surface?
m/s

Let's break down the problem and solve each part step by step. 

We are given the height function of the rock on Mars as:
\[
H(t) = 17t - 1.86t^2
\]
where \( H(t) \) is the height (in meters) and \( t \) is time (in seconds). 

### (a) Find the velocity after 2 seconds.
The velocity of the rock is the derivative of the height function with respect to time. So, we need to differentiate \( H(t) \):

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

Now, substitute \( t = 2 \) seconds into the velocity equation:

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

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

---

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

We already found the general formula for the velocity as:

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

For \( t = a \), we substitute \( t = a \) into this equation:

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

Thus, the velocity at time \( t = a \) is **\( 17 - 3.72a \, \text{m/s} \)**.

---

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

The rock will hit the surface when the height \( H(t) \) is zero. We need to solve for \( t \) when \( H(t) = 0 \):

\[
H(t) = 17t - 1.86t^2 = 0
\]

This is a quadratic equation:

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

Factor out \( t \):

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

This gives two solutions: \( t = 0 \) (the time when the rock was thrown) and \( 1.86t = 17 \). Solving for \( t \):

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

So, the rock will hit the surface after approximately **9.1 seconds**.

---

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

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

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

The negative sign indicates that the rock is moving downward. So, the velocity with which the rock hits the surface is approximately **-16.85 m/s**.

---

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

### Questions:
1. What would the height be at \( t = 3 \) seconds?
2. How high does the rock go at its peak?
3. What is the total time the rock spends in the air?
4. How does the height function change if we increase the initial velocity?
5. What would the velocity be after 5 seconds?

### Tip:
Remember, the velocity at any point in time is the rate of change of height, so by differentiating the height function, you get the velocity function!

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 of the rock after 2 seconds. (b) Find the velocity of the rock when t = a. (c) When will the rock hit the surface? (d) With what velocity will the rock hit the surface?

Explore the motion of a rock thrown upward on Mars with an initial velocity of 17 m/s. Learn how to find its velocity at different times, determine when it hits the ground, and calculate the impact speed using differentiation and quadratic equations. Suitable for students in Grades 10-12.

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