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 object's height after 10 seconds, we need to substitute \( x = 10 \) into the function \( y = -5(x - 4)^2 + 180 \).

### Step 1: Calculate the height at \( x = 10 \):

\[
y = -5(10 - 4)^2 + 180
\]
\[
y = -5(6)^2 + 180
\]
\[
y = -5(36) + 180
\]
\[
y = -180 + 180
\]
\[
y = 0
\]

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

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

The rate of change can be found by calculating the change in height over the change in time. We will first find the height at \( x = 5 \) and \( x = 8 \).

#### Height at \( x = 5 \):

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

#### Height at \( x = 8 \):

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

### Step 3: Calculate the rate of change:

Now we can find the rate of change from \( x = 5 \) to \( x = 8 \):

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

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