A person standing on the roof of a building throws a ball directly upward. The ball misses the rooftop on its

way down and eventually strikes the ground. The function s(t) = −16t2 + 64t + 80 describes the ball’s height
above the ground, s(t) , in feet, t , seconds after it was thrown.
a. Find the ball’s average velocity between the time it was thrown and 2 seconds later.
b. Find the ball’s average velocity between 2 and 4 seconds after it was thrown.

1 answer

Since v(t) = 64 - 32t

v(2) = 0

So, the velocity decreases from 64 to 0 in 2 seconds. Average v = 32

Similarly, once the ball starts falling, it regains its velocity of -64 after 2 seconds. So, on the way down, avg v = -32