Question

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
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. Please make it simple.

Answer 1

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

### Step 1: Calculate the height after 10 seconds
Substitute \( x = 10 \):
\[
y = -5(10 - 4)^2 + 180
\]
\[
y = -5(6)^2 + 180
\]
\[
y = -5(36) + 180
\]
\[
y = -180 + 180
\]
\[
y = 0
\]

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

### Step 2: Calculate the rate of change from 5 seconds to 8 seconds
To find the rate of change over the interval from 5 seconds to 8 seconds, we first need to find the heights at these two times.

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

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

#### Step 3: Calculate the rate of change
The rate of change between 5 seconds and 8 seconds can be calculated using the formula:
\[
\text{Rate of Change} = \frac{y(8) - y(5)}{8 - 5}
\]
Where \( y(5) = 175 \) and \( y(8) = 100 \):
\[
\text{Rate of Change} = \frac{100 - 175}{8 - 5}
\]
\[
\text{Rate of Change} = \frac{-75}{3}
\]
\[
\text{Rate of Change} = -25
\]

Thus, the rate of change of the object's height from 5 seconds to 8 seconds is **-25 meters per second**. This indicates that the object is descending at a rate of 25 meters per second during this interval.