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.

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 provided function \( y = -5(x - 4)^2 + 180 \):

1. **Calculate the height at \( x = 10 \)**:
\[
y = -5(10 - 4)^2 + 180
\]
\[
y = -5(6)^2 + 180
\]
\[
y = -5(36) + 180
\]
\[
y = -180 + 180
\]
\[
y = 0
\]

So, the object’s height after 10 seconds is \( \mathbf{0 \text{ meters}} \).

2. **Calculate the rate of change for the interval from 5 seconds to 8 seconds**:
First, find the height at \( x = 5 \) and \( x = 8 \).

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

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

Now we have the following heights:
- At \( x = 5 \): \( y = 175 \) meters
- At \( x = 8 \): \( y = 100 \) meters

Next, we find the rate of change of height over the interval from \( x = 5 \) to \( x = 8 \):
\[
\text{Rate of change} = \frac{y(8) - y(5)}{8 - 5}
\]
\[
= \frac{100 - 175}{8 - 5}
\]
\[
= \frac{-75}{3}
\]
\[
= -25 \text{ meters per second}
\]

Thus, the rate of change of the object's height from 5 seconds to 8 seconds is \( \mathbf{-25 \text{ meters per second}} \).