To find the height of the object after 10 seconds using the function \( y = -5(x - 4)^2 + 180 \), we will substitute \( x = 10 \) into the equation.
Step 1: Find the height after 10 seconds
\[ y = -5(10 - 4)^2 + 180 \]
Calculating inside the parentheses first:
\[ 10 - 4 = 6 \]
Now, substituting this value back into the equation:
\[ y = -5(6)^2 + 180 \]
Calculating \( 6^2 \):
\[ 6^2 = 36 \]
Now, substituting that back into the equation:
\[ y = -5(36) + 180 \]
Calculating \(-5 \times 36\):
\[ -5 \times 36 = -180 \]
Adding this to 180:
\[ y = -180 + 180 = 0 \]
Therefore, the object's height after 10 seconds is:
\[ \boxed{0 \text{ meters}} \]
Step 2: Find the rate of change from 5 seconds to 8 seconds
To find the rate of change, we need to calculate the heights at \( x = 5 \) and \( x = 8 \), and then use those values to determine the average rate of change over that interval.
Calculate the height at \( x = 5 \)
\[ y(5) = -5(5 - 4)^2 + 180 \] \[ = -5(1)^2 + 180 \] \[ = -5(1) + 180 \] \[ = -5 + 180 = 175 \]
So, the height at \( x = 5 \) seconds is:
\[ y(5) = 175 \text{ meters} \]
Calculate the height at \( x = 8 \)
\[ y(8) = -5(8 - 4)^2 + 180 \] \[ = -5(4)^2 + 180 \] \[ = -5(16) + 180 \] \[ = -80 + 180 = 100 \]
So, the height at \( x = 8 \) seconds is:
\[ y(8) = 100 \text{ meters} \]
Step 3: Calculate the rate of change
The rate of change of height \( y \) over the interval from 5 seconds to 8 seconds is:
\[ \text{Rate of change} = \frac{y(8) - y(5)}{8 - 5} \] \[ = \frac{100 - 175}{8 - 5} \] \[ = \frac{-75}{3} = -25 \]
Therefore, the rate of change of the object's height from 5 seconds to 8 seconds is:
\[ \boxed{-25 \text{ meters/second}} \]