Question

Q2 A. An object is launched from a platform. Its height, y (in meters), x seconds after the launch is modeled by the function
y=-5(x-4)2+180 What is the object’s height after 10 seconds?

Q2 B.What is the rate of change for the interval 5 seconds to 8 seconds? Show your work.

Answer 1

To solve the given problems, we'll first analyze the function provided:

\[ y = -5(x - 4)^2 + 180 \]

### Q2 A: Finding the height after 10 seconds.

To find the object's height after 10 seconds, we substitute \( x = 10 \) into the height equation:

\[
y = -5(10 - 4)^2 + 180
\]

Calculating inside the parentheses:

\[
y = -5(6)^2 + 180
\]

Calculating the square:

\[
y = -5(36) + 180
\]

Now performing the multiplication:

\[
y = -180 + 180
\]

Finally, simplifying gives us:

\[
y = 0
\]

So, the object's height after 10 seconds is **0 meters**.

### Q2 B: Finding the rate of change for the interval from 5 seconds to 8 seconds.

The rate of change can be calculated using the formula:

\[
\text{Rate of Change} = \frac{y(t_2) - y(t_1)}{t_2 - t_1}
\]

where \( t_1 = 5 \) seconds and \( t_2 = 8 \) seconds.

**Step 1: Find \( y(5) \)**

\[
y(5) = -5(5 - 4)^2 + 180
\]
\[
y(5) = -5(1)^2 + 180
\]
\[
y(5) = -5 + 180
\]
\[
y(5) = 175
\]

**Step 2: Find \( y(8) \)**

\[
y(8) = -5(8 - 4)^2 + 180
\]
\[
y(8) = -5(4)^2 + 180
\]
\[
y(8) = -5(16) + 180
\]
\[
y(8) = -80 + 180
\]
\[
y(8) = 100
\]

**Step 3: Calculate the rate of change.**

Now we substitute \( y(5) \) and \( y(8) \) into the rate of change formula:

\[
\text{Rate of Change} = \frac{y(8) - y(5)}{8 - 5} = \frac{100 - 175}{8 - 5} = \frac{-75}{3} = -25
\]

Thus, the rate of change of the height of the object from 5 seconds to 8 seconds is **-25 meters per second**.