Question

An object is launched from a platform it's height y in meters X x for seconds after the launch is modeled by the function y = -5 (x - 4) squared + 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 given height function:

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

we substitute \(x = 10\) into the equation:

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

Calculating this step by step:

1. Calculate \(10 - 4\):
\[
10 - 4 = 6
\]

2. Find \((6)^2\):
\[
6^2 = 36
\]

3. Multiply by \(-5\):
\[
-5 \times 36 = -180
\]

4. Add 180:
\[
y = -180 + 180 = 0
\]

So, the object's height after 10 seconds is \(0\) meters.

Next, we need to find the rate of change of height over the interval from 5 seconds to 8 seconds. To do this, we will evaluate the height function at \(x = 5\) and \(x = 8\).

**Finding height at \(x = 5\):**

\[
y(5) = -5(5 - 4)^2 + 180
\]
Calculating:

1. Calculate \(5 - 4\):
\[
5 - 4 = 1
\]

2. Find \((1)^2\):
\[
1^2 = 1
\]

3. Multiply by \(-5\):
\[
-5 \times 1 = -5
\]

4. Add 180:
\[
y(5) = -5 + 180 = 175
\]

**Finding height at \(x = 8\):**

\[
y(8) = -5(8 - 4)^2 + 180
\]
Calculating:

1. Calculate \(8 - 4\):
\[
8 - 4 = 4
\]

2. Find \((4)^2\):
\[
4^2 = 16
\]

3. Multiply by \(-5\):
\[
-5 \times 16 = -80
\]

4. Add 180:
\[
y(8) = -80 + 180 = 100
\]

**Now, we can calculate the rate of change over the interval from 5 to 8 seconds:**

Rate of change is given by:

\[
\text{Rate of change} = \frac{y(8) - y(5)}{8 - 5}
\]

Substituting the heights we found:

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

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