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. 1 sentence

Answer 1

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

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

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

Next, to find the rate of change of height from 5 seconds to 8 seconds, we calculate the height at both times and then find the difference over the time interval:

1. Calculate height at \( x = 5 \):
\[
y(5) = -5(5 - 4)^2 + 180
\]
\[
y(5) = -5(1)^2 + 180
\]
\[
y(5) = -5 + 180 = 175
\]

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

3. Now, compute the rate of change:
\[
\text{Rate of Change} = \frac{y(8) - y(5)}{8 - 5} = \frac{100 - 175}{3} = \frac{-75}{3} = -25
\]

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