f a rock is thrown vertically upward from the surface of Mars with velocity of 25m/s, its height (in meters) after *t* seconds is 

*h* =25*t* − 1.86*t*2.

(a)

What is the velocity (in m/s) of the rock after2s?

m/s

(b)

What is the velocity (in m/s) of the rock when its height is45m on its way up? On its way down? (Round your answers to two decimal places.)

upm/s downm/s

To solve this problem, we will first derive the velocity function from the given height equation, then use it to find the velocity at specific times and conditions.

### Given:
\[
h(t) = 25t - 1.86t^2
\]

### Part (a): Velocity after 2 seconds

To find the velocity, we need to take the derivative of the height function \( h(t) \) with respect to \( t \), which gives us the velocity function \( v(t) \):

\[
v(t) = \frac{d}{dt}(25t - 1.86t^2)
\]

Calculating the derivative:

\[
v(t) = 25 - 3.72t
\]

Now, substitute \( t = 2 \) to find the velocity after 2 seconds:

\[
v(2) = 25 - 3.72 \times 2
\]
\[
v(2) = 25 - 7.44 = 17.56 \, \text{m/s}
\]

### Answer (a): 
The velocity after 2 seconds is **17.56 m/s**.

---

### Part (b): Velocity when height is 45 m

We need to find the times when the height of the rock is 45 meters. So, we set \( h(t) = 45 \) and solve for \( t \):

\[
45 = 25t - 1.86t^2
\]

Rearranging this into a standard quadratic form:

\[
1.86t^2 - 25t + 45 = 0
\]

This is a quadratic equation in the form \( at^2 + bt + c = 0 \), where:
- \( a = 1.86 \)
- \( b = -25 \)
- \( c = 45 \)

Use the quadratic formula to solve for \( t \):

\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Substitute the values:

\[
t = \frac{-(-25) \pm \sqrt{(-25)^2 - 4 \times 1.86 \times 45}}{2 \times 1.86}
\]
\[
t = \frac{25 \pm \sqrt{625 - 334.8}}{3.72}
\]
\[
t = \frac{25 \pm \sqrt{290.2}}{3.72}
\]
\[
t = \frac{25 \pm 17.03}{3.72}
\]

Now, calculate the two possible values of \( t \):

\[
t_1 = \frac{25 + 17.03}{3.72} \approx 11.3 \, \text{seconds (on the way down)}
\]
\[
t_2 = \frac{25 - 17.03}{3.72} \approx 2.14 \, \text{seconds (on the way up)}
\]

### Finding velocities at these times:

1. **Velocity when the rock is on the way up (\( t_2 = 2.14 \))**:

\[
v(2.14) = 25 - 3.72 \times 2.14
\]
\[
v(2.14) \approx 25 - 7.96 = 17.04 \, \text{m/s}
\]

2. **Velocity when the rock is on the way down (\( t_1 = 11.3 \))**:

\[
v(11.3) = 25 - 3.72 \times 11.3
\]
\[
v(11.3) \approx 25 - 42.04 = -17.04 \, \text{m/s}
\]

The negative sign indicates that the rock is moving downward at that time.

### Answer (b): 
- Velocity when height is 45 m on its way up: **17.04 m/s**.
- Velocity when height is 45 m on its way down: **-17.04 m/s**.

---

Would you like more details or have any questions?

Here are 5 related questions:
1. What is the maximum height reached by the rock?
2. How long does it take for the rock to reach its maximum height?
3. At what time does the rock hit the ground again?
4. What is the initial acceleration of the rock?
5. How would the height formula change if the gravity on Mars were different?

**Tip:** When dealing with projectile motion, using derivatives to find velocity and acceleration is a powerful approach. The first derivative of position gives velocity, and the second derivative gives acceleration.

If a rock is thrown vertically upward from the surface of Mars with a velocity of 25 m/s, its height (in meters) after t seconds is h = 25t − 1.86t^2.

(a) What is the velocity (in m/s) of the rock after 2 seconds?

(b) What is the velocity (in m/s) of the rock when its height is 45 m on its way up? On its way down?

Explore projectile motion by calculating the velocity of a rock thrown vertically upward from Mars with an initial speed of 25 m/s. Find the velocity after 2 seconds, and determine the upward and downward velocities when the rock reaches 45 m. Concepts include differentiation and quadratic equations, suitable for high school physics and calculus students.

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