h () = ANSWER: The height of the ball at given initial snapshot of time t = 0.75 second.

initial height of ball at h(t = 0) is 4 feet. How do you know it is 4 feet? You refer to the given information of the height equation; h(t) = - 16 t ² + 20 t + 4 . By substituting in equation t = 0, you will get height , h(t=0) = 4 feet.
Imagine you measure the height from a person's hand holding the ball to the ground before he throws it upward.

and final time, t _final =

h ( ) = ² + +

h ( ) = + +

h ( ) = final height of the ball at final given time, t = 1.25 second

At time = 0.75 seconds, the object is still at the air, the height from the ground is 10 feet.

At time = 1.25 seconds, the object is falling down, the height from the ground is 4 feet

Average Rate of Change

Given the function:
f(t) = -16t² + 20t + 4

We want the average rate of change on the interval [0.75, 1.25], which is:
(f(1.25) − f(0.75)) / (1.25 − 0.75)

Step 1: Compute f(1.25)

f(1.25) = -16(1.25²) + 20(1.25) + 4
1.25² = 1.5625
f(1.25) = -16(1.5625) + 25 + 4
f(1.25) = -25 + 29 = 4

Step 2: Compute f(0.75)

f(0.75) = -16(0.75²) + 20(0.75) + 4
0.75² = 0.5625
f(0.75) = -16(0.5625) + 15 + 4
f(0.75) = -9 + 19 = 10

Step 3: Average rate of change

Average rate of change = (f(1.25) − f(0.75)) / (1.25 − 0.75)
= (4 − 10) / (1.25 − 0.75)
= -6 / 0.5
= -12

Final Answer

The average rate of change over [0.75, 1.25] is -12 feet per second.