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. 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.