To solve the given problems, we'll first analyze the function provided:
\[ y = -5(x - 4)^2 + 180 \]
Q2 A: Finding the height after 10 seconds.
To find the object's height after 10 seconds, we substitute \( x = 10 \) into the height equation:
\[ y = -5(10 - 4)^2 + 180 \]
Calculating inside the parentheses:
\[ y = -5(6)^2 + 180 \]
Calculating the square:
\[ y = -5(36) + 180 \]
Now performing the multiplication:
\[ y = -180 + 180 \]
Finally, simplifying gives us:
\[ y = 0 \]
So, the object's height after 10 seconds is 0 meters.
Q2 B: Finding the rate of change for the interval from 5 seconds to 8 seconds.
The rate of change can be calculated using the formula:
\[ \text{Rate of Change} = \frac{y(t_2) - y(t_1)}{t_2 - t_1} \]
where \( t_1 = 5 \) seconds and \( t_2 = 8 \) seconds.
Step 1: Find \( y(5) \)
\[ y(5) = -5(5 - 4)^2 + 180 \] \[ y(5) = -5(1)^2 + 180 \] \[ y(5) = -5 + 180 \] \[ y(5) = 175 \]
Step 2: Find \( y(8) \)
\[ y(8) = -5(8 - 4)^2 + 180 \] \[ y(8) = -5(4)^2 + 180 \] \[ y(8) = -5(16) + 180 \] \[ y(8) = -80 + 180 \] \[ y(8) = 100 \]
Step 3: Calculate the rate of change.
Now we substitute \( y(5) \) and \( y(8) \) into the rate of change formula:
\[ \text{Rate of Change} = \frac{y(8) - y(5)}{8 - 5} = \frac{100 - 175}{8 - 5} = \frac{-75}{3} = -25 \]
Thus, the rate of change of the height of the object from 5 seconds to 8 seconds is -25 meters per second.