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?

Answer 1

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

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

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

Next, we will calculate the rate of change of height for the interval from 5 seconds to 8 seconds. The rate of change can be found using the formula:

\[
\text{Rate of change} = \frac{y(t_2) - y(t_1)}{t_2 - t_1}
\]

where \( t_1 = 5 \) seconds and \( t_2 = 8 \) seconds.

First, we find \( y(5) \):

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

Next, we find \( y(8) \):

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

Now, substituting these values into the rate of change formula:

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

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

To summarize:
- The object's height after 10 seconds is **0 meters**.
- The rate of change from 5 seconds to 8 seconds is **-25 meters per second**.