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

Question

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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.