Question

Q2 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?




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

Answer 1

To find the height of the object after 10 seconds using the function \( y = -5(x - 4)^2 + 180 \), we will substitute \( x = 10 \) into the equation.

### Step 1: Find the height after 10 seconds

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

Calculating inside the parentheses first:

\[
10 - 4 = 6
\]

Now, substituting this value back into the equation:

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

Calculating \( 6^2 \):

\[
6^2 = 36
\]

Now, substituting that back into the equation:

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

Calculating \(-5 \times 36\):

\[
-5 \times 36 = -180
\]

Adding this to 180:

\[
y = -180 + 180 = 0
\]

Therefore, the object's height after 10 seconds is:

\[
\boxed{0 \text{ meters}}
\]

### Step 2: Find the rate of change from 5 seconds to 8 seconds

To find the rate of change, we need to calculate the heights at \( x = 5 \) and \( x = 8 \), and then use those values to determine the average rate of change over that interval.

#### Calculate the height at \( x = 5 \)

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

So, the height at \( x = 5 \) seconds is:

\[
y(5) = 175 \text{ meters}
\]

#### Calculate the height at \( x = 8 \)

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

So, the height at \( x = 8 \) seconds is:

\[
y(8) = 100 \text{ meters}
\]

### Step 3: Calculate the rate of change

The rate of change of height \( y \) over the interval from 5 seconds to 8 seconds is:

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

Therefore, the rate of change of the object's height from 5 seconds to 8 seconds is:

\[
\boxed{-25 \text{ meters/second}}
\]