An object attached to a spring bounces up and down. Its height is given by

*h*(*t*) = 60 − 20cos(0.2*t* + 0.4)

where *h* is in inches and *t* is in seconds.

Answer the following questions accurate to two decimal digits and include correct units.
 
Find the height of the object at the second instant in time *t* > 0 its velocity is 2 in/s

We are given the height function \( h(t) = 60 - 20 \cos(0.2t + 0.4) \), where \( h(t) \) represents the height of the object in inches at time \( t \) seconds.

To solve the problem, we need to find the second time instant \( t \) when the velocity of the object is 2 inches per second. The velocity is the derivative of the height function with respect to time \( t \), so we start by finding \( h'(t) \), the velocity function.

### Step 1: Compute the derivative of \( h(t) \)
The height function is:
\[
h(t) = 60 - 20 \cos(0.2t + 0.4)
\]
The derivative with respect to \( t \) is:
\[
h'(t) = 20 \sin(0.2t + 0.4) \cdot \frac{d}{dt}(0.2t + 0.4)
\]
Since the derivative of \( 0.2t + 0.4 \) is 0.2, we have:
\[
h'(t) = 20 \cdot \sin(0.2t + 0.4) \cdot 0.2
\]
\[
h'(t) = 4 \sin(0.2t + 0.4)
\]

### Step 2: Set the velocity equal to 2 inches per second
We want to find the second time instant when \( h'(t) = 2 \). Therefore, we solve:
\[
4 \sin(0.2t + 0.4) = 2
\]
Divide both sides by 4:
\[
\sin(0.2t + 0.4) = \frac{1}{2}
\]

### Step 3: Solve for \( t \)
The general solution for \( \sin x = \frac{1}{2} \) occurs when:
\[
x = \frac{\pi}{6} + 2k\pi \quad \text{or} \quad x = \pi - \frac{\pi}{6} + 2k\pi \quad \text{for integer } k
\]
Simplifying gives:
\[
x = \frac{\pi}{6} + 2k\pi \quad \text{or} \quad x = \frac{5\pi}{6} + 2k\pi
\]

Now, substitute \( x = 0.2t + 0.4 \) into this equation:
\[
0.2t + 0.4 = \frac{\pi}{6} + 2k\pi \quad \text{or} \quad 0.2t + 0.4 = \frac{5\pi}{6} + 2k\pi
\]

Solving for \( t \), we get two cases.

#### Case 1: \( 0.2t + 0.4 = \frac{\pi}{6} \)
\[
0.2t = \frac{\pi}{6} - 0.4
\]
\[
t = \frac{\frac{\pi}{6} - 0.4}{0.2}
\]

#### Case 2: \( 0.2t + 0.4 = \frac{5\pi}{6} \)
\[
t = \frac{\frac{5\pi}{6} - 0.4}{0.2}
\]

Let me calculate the exact value of \( t \).The two possible time instants are:

- \( t_1 = 0.62 \) seconds (first time when velocity is 2 in/s)
- \( t_2 = 11.09 \) seconds (second time when velocity is 2 in/s)

Since we are asked for the second instant, we will use \( t_2 = 11.09 \) seconds.

### Step 4: Find the height at \( t = 11.09 \)
To find the height at this time, substitute \( t_2 = 11.09 \) into the height equation:
\[
h(t) = 60 - 20 \cos(0.2t + 0.4)
\]

Let's compute this value.It seems like I can’t do more advanced data analysis right now. Please try again later. 

However, you can easily calculate the height at \( t = 11.09 \) seconds by substituting the value into the height equation:
\[
h(t) = 60 - 20 \cos(0.2t + 0.4)
\]
where \( t = 11.09 \).

Let me know if you need further assistance or details! Here are some related questions:

1. What is the maximum and minimum height of the object?
2. How long does it take for the object to complete one full bounce?
3. What is the velocity of the object at \( t = 5 \) seconds?
4. Can you find the time when the object reaches its maximum height?
5. What is the acceleration of the object at \( t = 11.09 \) seconds?

**Tip**: The cosine function in height equations like this one describes oscillations, so the object’s motion is periodic.

An object attached to a spring bounces up and down. Its height is given by h(t) = 60 − 20cos(0.2t + 0.4), where h is in inches and t is in seconds. Find the height of the object at the second instant in time t > 0 its velocity is 2 in/s.

Learn how to determine the second time an object in harmonic motion has a velocity of 2 inches per second, with its height described by h(t) = 60 - 20cos(0.2t + 0.4). This problem covers trigonometric functions and derivatives, ideal for Grades 11-12.

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