Question

An object is launched from a platform it's height y and meters X in seconds after the launch is modeled by the function y = -5(x - 4) squared +180 what is the object's height after 10 seconds what is the rate of change for the interval 5 seconds to 8 seconds

Answer 1

To solve the problem, we'll first evaluate the object's height after 10 seconds and then find the rate of change between 5 seconds and 8 seconds.

### Step 1: Find the object's height after 10 seconds.
We use the function provided:
\[
y = -5(x - 4)^2 + 180
\]
Substituting \(x = 10\):
\[
y = -5(10 - 4)^2 + 180
\]
\[
y = -5(6)^2 + 180
\]
\[
y = -5(36) + 180
\]
\[
y = -180 + 180
\]
\[
y = 0
\]

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

### Step 2: Find the rate of change from 5 seconds to 8 seconds.
To find the rate of change, we need to evaluate the height at \(x = 5\) and \(x = 8\).

#### Height at \(x = 5\):
\[
y = -5(5 - 4)^2 + 180
\]
\[
= -5(1)^2 + 180
\]
\[
= -5 + 180
\]
\[
= 175
\]

#### Height at \(x = 8\):
\[
y = -5(8 - 4)^2 + 180
\]
\[
= -5(4)^2 + 180
\]
\[
= -5(16) + 180
\]
\[
= -80 + 180
\]
\[
= 100
\]

#### Calculate the rate of change:
The rate of change is given by the difference in height divided by the difference in time:
\[
\text{Rate of change} = \frac{y(8) - y(5)}{8 - 5}
\]
\[
= \frac{100 - 175}{8 - 5}
\]
\[
= \frac{-75}{3}
\]
\[
= -25
\]

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